A Beginner's Guide to Differential Forms

Peter Collier

Incomprehensible Books
For Anne

Contents

Preface ..... 11
1 Preliminaries ..... 15
1.1 Manifolds ..... 15
1.2 Vectors ..... 17
1.3 Tangent vectors ..... 18
1.4 Parameterisation ..... 19
1.5 Notation for forms ..... 20
1.6 Cross product ..... 20
1.7 Determinants ..... 21
1.7.1 The determinant of a 1 × 1 1 × 1 1xx11 \times 11×1 matrix ..... 21
1.7.2 The determinant of a 2 × 2 2 × 2 2xx22 \times 22×2 matrix ..... 22
1.7.3 The determinant of a 3 × 3 3 × 3 3xx33 \times 33×3 matrix ..... 23
1.7.4 The determinate of an n × n n × n n xx nn \times nn×n matrix, where n > 3 n > 3 n > 3n>3n>3 ..... 24
2 What are differential forms? ..... 25
2.10 -forms ..... 26
2.2 1-forms ..... 26
2.2.1 The meaning of d x i d x i dx^(i)d x^{i}dxi ..... 28
2.2.2 How this works - basis 1-forms and basis vectors ..... 28
2.2.3 d x i d x i dx^(i)d x^{i}dxi picks out a vector's i i iii th component ..... 30
2.2.4 An example using polar coordinates ..... 31
2.2.4.1 Calculating the basis vectors and 1-forms ..... 31
2.2.4.2 Tangent vectors to a circle ..... 33
2.2.5 Curves ..... 34
2.2.6 The differential d f d f dfd fdf revisited ..... 35
2.2.6.1 d f ( v ) d f ( v ) df(v)d f(\mathbf{v})df(v) as an approximation to Δ f Δ f Delta f\Delta fΔf ..... 36
2.2.6.2 Vectors as differential operators ..... 38
2.2.7 Another 1-form example ..... 41
2.3 -forms ..... 42
2.3.1 The wedge product ..... 43
2.3.2 2-forms acting on two vectors ..... 46
2.3.3 2 -form example ..... 49
2.3.4 Surfaces ..... 49
2.43 -forms and higher ..... 50
2.4.1 3 -forms acting on three vectors ..... 51
2.4.2 3 -form example ..... 54
2.4.3 Three and higher dimensional spaces ..... 54
3 Converting between differential forms and vectors ..... 57
3.10 -forms ..... 57
3.2 1-forms ..... 57
3.3 2-forms ..... 58
3.4 3-forms ..... 59
4 Differentiation ..... 61
5 Div, grad and curl ..... 65
5.1 Gradient and 0 -forms ..... 65
5.2 Curl and 1-forms ..... 65
5.3 Divergence and 2 -forms ..... 67
5.4 A couple of vector identities ..... 68
6 Orientation ..... 69
7 Integrating differential forms ..... 75
7.1 Change of variables ..... 76
7.1.1 The pullback ..... 79
7.2 Defining the integral ..... 82
7.2.1 k k kkk-form integrated over a k k kkk-dimensional manifold in R n R n R^(n)\mathbb{R}^{n}Rn, where
k < n k < n k < nk<nk<n ..... 82
7.2.2 n n nnn-form integrated over an n n nnn-dimensional manifold in R n R n R^(n)\mathbb{R}^{n}Rn ..... 84
7.2.3 The recipe ..... 86
7.2.4 Some examples ..... 86
7.3 Independence of parameterisation ..... 92
7.3.1 Parameterised curves ..... 92
7.3.2 Parameterised surfaces ..... 93
8 Integrating differential forms and vector calculus ..... 95
8.1 Line integrals ..... 96
8.1.1 Conservative vector fields ..... 98
8.1.2 And with differential forms ..... 99
8.2 Surface integrals ..... 100
8.2.1 And with differential forms ..... 103
8.3 Volume integrals ..... 105
8.3.1 And with differential forms ..... 106
9 The generalised Stokes' theorem ..... 107
9.1 The gradient theorem ..... 108
9.1.1 Generalised Stokes' theorem ..... 109
9.2 Green's theorem ..... 110
9.2.1 Generalised Stokes' theorem ..... 111
9.3 Stokes' theorem ..... 112
9.3.1 Generalised Stokes' theorem ..... 113
9.4 Divergence theorem ..... 114
9.4.1 Generalised Stokes' theorem ..... 115
10 Maxwell's equations ..... 117
10.1 The vector calculus approach ..... 117
10.2 The differential forms approach ..... 120
10.2.1 The Hodge star operator ..... 120
10.2.2 Moving on ..... 121
10.2.3 Gauss's law for magnetism and Faraday's law ..... 122
10.2.4 Gauss's law and the Ampère-Maxwell law ..... 124
11 Three nice results from topology ..... 127
11.1 The drum theorem ..... 131
11.2 The Brouwer fixed-point theorem ..... 133
11.3 The hairy ball theorem. ..... 135
11.3.1 Proof of Equation (11.3.2) ..... 139
Bibliography ..... 143
Index ..... 145

Preface

There is an intriguing class of mathematical objects, called differential forms, that live on manifolds, eat tangent vectors and spit out numbers, and do this in a way that makes them of great interest to mathematicians and physicists. Arfken et al [3], authors of Mathematical Methods for Physicists: A Comprehensive Guide, comment:
The calculus of differential forms, of which the leading developer was Elie Cartan, has become recognized as a natural and very powerful tool for the treatment of curved coordinates, both in classical settings and in contemporary studies of curved spacetime. Cartan's calculus leads to a remarkable unification of concepts and theorems of vector analysis that is worth pursuing, with the result that in differential geometry and in theoretical physics the use of differential forms is now widespread.
And this is from Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach by Hubbard and Hubbard [11]:
Because forms work in any dimension, they are the natural way to approach two towering subjects that are inherently four-dimensional: electromagnetism and the theory of relativity. They also provide a unified treatment of differentiation and of the fundamental theorem of calculus: one operator (the exterior derivative) works in all dimensions, and one short, elegant statement (the generalized Stokes's theorem) generalizes the fundamental theorem of calculus to all dimensions. In contrast, vector calculus requires special formulas, operators, and theorems for each dimension where it works.
I first came across differential forms some years ago when I was taking my first baby steps in studying relativity theory 1 1 ^(1){ }^{1}1. An infinitesimal displacement in spacetime can be represented either as a tangent vector or a 1-form, dual objects linked by a function called the metric tensor. Not having much in the way of a mathematical imagination, I found there to be a whiff of magic to these mysterious 1 -forms that were part of the machinery used to describe the structure of curved spaces. And if 1-forms were perplexing, what on earth was the story behind their weightier relatives -2 -forms, 3 -forms, n n nnn-forms? Fast forward a few years, and with the differential forms bee still buzzing around in my head, I thought it would be fun to investigate these strange objects further.
I'm not a mathematician, however, and I found much of the mathematical literature to be so overwhelming and abstract as to be useless for the purposes of self-study. The
'Intrinsic definition' of differential forms given in the Wikipedia article, for example, includes stuff like β p ∈ Λ k T p ⋆ M β p ∈ Λ k T p ⋆ M beta_(p)inLambda^(k)T_(p)^(***)M\beta_{p} \in \Lambda^{k} T_{p}^{\star} Mβp∈ΛkTp⋆M and β p : ⨁ n = 1 k T p M → R β p : ⨁ n = 1 k   T p M → R beta_(p):bigoplus_(n=1)^(k)T_(p)M rarrR\beta_{p}: \bigoplus_{n=1}^{k} T_{p} M \rightarrow \mathbb{R}βp:⨁n=1kTpM→R, both of which I'm sure are crystal clear to a mathematics graduate but are way above my head.
So not being a mathematician, I knew I wouldn't be able to master differential forms to an advanced level. For the foreseeable future at least, β p : ⨁ n = 1 k T p M → R β p : ⨁ n = 1 k   T p M → R beta_(p):bigoplus_(n=1)^(k)T_(p)M rarrR\beta_{p}: \bigoplus_{n=1}^{k} T_{p} M \rightarrow \mathbb{R}βp:⨁n=1kTpM→R would remain a mystery to me. However, I'm a firm believer that given sufficient enthusiasm, a little maths education can go a long way. With that optimistic philosophy in mind, my goal was to achieve a basic working understanding of the subject matter - what are differential forms? how are they manipulated? what are they used for? In the hope that others might share my interest, the notes I put together during my studies eventually morphed into this little book, which is aimed at those readers who may not be maths whizzes but who are interested in tackling a relaxed but wide-ranging introduction to differential forms.
The only prerequisite - apart from the aforementioned enthusiasm - is a reasonable foundation in advanced, school-level mathematics. So ideally you should be familiar with basic algebra and sufficient multivariable calculus to be at ease with partial derivatives and double and triple integrals. Some vector analysis would also be useful, as would a little bit of matrix algebra. We make many references to manifolds, but mostly only at the most elementary level as smooth curves, surfaces and higher dimensional spaces embedded in some Euclidean R n R n R^(n)\mathbb{R}^{n}Rn. Because my target audience is only a small maths-step up from the general reader, I've attempted to make verbosity a virtue, giving what I consider to be important derivations in full, even at the risk of stating and restating what is blindingly obvious to the more mathematically astute.
The emphasis for much of this book will be on how differential forms provide an alternative means of understanding three-dimensional vector calculus. For those readers unfamiliar with the basics of vector calculus I summarise the key concepts as we encounter them. Worked examples are a great way of consolidating mathematical understanding, so I've included a number of such problems in the text. Where possible, I've included code that you can copy into the WolframAlpha online calculator [22] to check your answers.
We start with a short review of essential background material and notation. Next we look at differential forms and what we mean when we say they are linear or, more generally, multilinear alternating functions of tangent vectors. We then look at how differential forms correspond with various scalar and vector fields; how to differentiate forms; and how to express in the language of differential forms the three important operators of vector analysis - grad, curl and div. Differential forms are integrated over oriented manifolds, so we then briefly discuss the notion of orientation. Differential forms are the answer to the question: what objects do we integrate on manifolds? 1 -forms are the natural things to integrate along a curve, 2 -forms over a surface, and so on. Therefore, we next move on to integrating differential forms, in the process seeing how forms can be used to easily derive the change of variables formula. We then relate what we have learned to the line, surface and volume integrals of vector calculus.
These integrals are themselves the building blocks of the big four vector calculus theorems: the gradient theorem, Green's theorem, Stokes' theorem and the divergence theorem. In turn, these four theorems are themselves special cases of the generalised Stokes' theorem - what Schulz and Schulz [18] refers to as 'one of the triumphs of elementary mathematics' - that applies to spaces of arbitrary dimension and which we discuss in chapter 9. Next we turn to Maxwell's equations, the foundation of classical electromagnetic theory. We discuss what these equations are, what they mean, and how, using differential forms, they can be pleasingly reduced to just two concise formulations. Finally, we see how the generalised Stokes' theorem can be used to prove three rather neat topological theorems: the drum theorem, the Brouwer fixed-point theorem and the famous hairy ball theorem ('you can't comb a hairy ball smooth').

Acknowledgements

I am greatly indebted to David L. Finn (Associate Professor of Mathematics at RoseHulman Institute of Technology), who was kind enough to read through my manuscript and provide invaluable comments and feedback. As ever, of course, any remaining errors are my own.
  • Comments and suggestions? Email the author at incomprehensiblething@gmail.com.
  • This book was written using L Y X L Y X L_(Y)X\mathrm{L}_{Y} \mathrm{X}LYX, an excellent maths-friendly (and much more) open source document preparation system based on LT E X LT E X LT_(E)X\mathrm{LT}_{\mathrm{E}} \mathrm{X}LTEX.
My apologies if any acknowledgement or bibliographic citation has been inadvertently omitted. Please contact me and I will be pleased to make the necessary arrangements at the earliest opportunity.
All images copyright @ Peter Collier 2021 except for:
Chapter 11: (1) Baby hairy head, author - Nojhan, licensed under the Creative Commons Attribution-Share Alike 2.5 Generic, 2.0 Generic and 1.0 Generic license, URL https://commons.wikimedia.org/wiki/File:Baby_hairy_head_DSCN2483.jpg. (2) Mona Lisa, by Leonardo da Vinci, from C2RMF retouched, public domain, URL https : //commons.wikimedia.org/wiki/File:Mona_Lisa,_by_Leonardo_da_Vinci,from C2RMF_retouched.jpg.

1 Preliminaries

Before we can begin our discussion of differential forms, we need to run through some essential background material and notation

1.1 Manifolds

Differential forms live on manifolds. So, what's a manifold?
Answer: as far as we're concerned, they're simply smooth (ie differentiable) spaces that are locally Euclidean. The surface of the Earth, for example, is not Euclidean (for the sake of argument we'll ignore the bumps and dips and assume our planet is perfectly spherical). Lines of longitude that start off parallel meet at the poles; the sum of the interior angle of a triangle need not add up to 180 ∘ 180 ∘ 180^(@)180^{\circ}180∘; and the Pythagorean theorem does not hold. However, a small enough patch of a sphere's surface is pretty much flat, ie Euclidean. (Mathematicians are able to formalise the notion of 'pretty much flat'.) The sphere is a two-dimensional manifold. Why two-dimensional? Because any point on the surface of a sphere can be described using a minimum of two coordinates - longitude and latitude, for example.
Circles and smooth (non-crossing) curves are examples of one-dimensional manifolds small enough segments look like straight lines and points on them can be described using a single coordinate.
One of the simplest manifolds is R n R n R^(n)\mathbb{R}^{n}Rn, which denotes n n nnn-dimensional Euclidean space. R 2 R 2 R^(2)\mathbb{R}^{2}R2 is two-dimensional, like a flat piece of paper. R 3 R 3 R^(3)\mathbb{R}^{3}R3 is three-dimensional, like the space we live in. Higher dimensional spaces ( n > 3 ) ( n > 3 ) (n > 3)(n>3)(n>3) are perfectly acceptable though hard for most of us to visualise.
Draw a smooth curve on a flat piece of paper and you have embedded a one-dimensional manifold (the curve) in two-dimensional Euclidean space R 2 R 2 R^(2)\mathbb{R}^{2}R2 (the flat piece of paper). A smoothly curved piece of zero-thickness wire lying on a table represents a onedimensional manifold (the wire) embedded in our own everyday three-dimensional Euclidean space R 3 R 3 R^(3)\mathbb{R}^{3}R3. A ball sitting next to the wire represents a two-dimensional manifold (the surface of the ball) embedded in R 3 R 3 R^(3)\mathbb{R}^{3}R3. Embedded manifolds such as these are nice and easy to visualise and are the ones we'll mainly be looking at. However, manifolds don't have to be embedded in some higher dimensional space and, though more abstract to describe, are perfectly capable of existing in their own right.
In relativity theory, spacetime is a four-dimensional (non-Euclidean) manifold - three of space, one of time.
We'll only be concerned with orientable manifolds - in chapter 6 we discuss what it means for a manifold to be orientable.
Here are some of the convention's we'll be making use of:
  • The letter C C CCC denotes a smooth curve, which can exist in any dimension.
  • The letter S S SSS denotes a two-dimensional smooth surface in R n R n R^(n)\mathbb{R}^{n}Rn. The unit disk on the x y x y xyx yxy plane in R 2 R 2 R^(2)\mathbb{R}^{2}R2, for example, consisting of all the points whose distance from the origin is less than or equal to 1 . Or a parameterised surface such as the top half of a unit radius sphere in R 3 R 3 R^(3)\mathbb{R}^{3}R3.
  • The symbol ∂ ∂ del\partial∂ (when not being used for partial derivatives) denotes a boundary. So the boundary of a surface S S SSS (if it has one) is ∂ S ∂ S del S\partial S∂S. The boundary of an n n nnn-dimensional manifold, assuming one exists, is itself a manifold of dimension n − 1 n − 1 n-1n-1n−1. For example, the boundary ∂ S ∂ S del S\partial S∂S of the top half of a (two dimensional) unit radius sphere S S SSS is a (one dimensional) unit radius circle.
  • The letter M M MMM denotes a space of any dimension (not just one, two or three). The boundary of M M MMM (if it has one) is ∂ M ∂ M del M\partial M∂M. So if M M MMM is a solid ball in R 3 R 3 R^(3)\mathbb{R}^{3}R3, its boundary, ∂ M ∂ M del M\partial M∂M, is a sphere.
We'll be meeting other types of manifold as we progress, but for now that's enough to be going on with.
A quick word about coordinates. We'll mainly be using Cartesian or rectangular coordinates x , y , z x , y , z x,y,zx, y, zx,y,z to describe R 3 R 3 R^(3)\mathbb{R}^{3}R3. Sometimes we'll use general coordinates u , v , w u , v , w u,v,wu, v, wu,v,w or u 1 , u 2 , u 3 , … , u n u 1 , u 2 , u 3 , … , u n u^(1),u^(2),u^(3),dots,u^(n)u^{1}, u^{2}, u^{3}, \ldots, u^{n}u1,u2,u3,…,un or x 1 , x 2 , x 3 , … , x n x 1 , x 2 , x 3 , … , x n x^(1),x^(2),x^(3),dots,x^(n)x^{1}, x^{2}, x^{3}, \ldots, x^{n}x1,x2,x3,…,xn. To avoid confusion - the context should make the meaning clear - remember the superscripts 1 , 2 , 3 1 , 2 , 3 1,2,31,2,31,2,3, etc are coordinate indices, not exponents, so x 2 x 2 x^(2)x^{2}x2 represents the second coordinate, not x x xxx squared. For example, using spherical coordinates r , θ , ϕ r , θ , ϕ r,theta,phir, \theta, \phir,θ,ϕ we would write
x 1 = r x 2 = θ x 3 = ϕ x 1 = r x 2 = θ x 3 = ϕ {:[x^(1)=r],[x^(2)=theta],[x^(3)=phi]:}\begin{aligned} & x^{1}=r \\ & x^{2}=\theta \\ & x^{3}=\phi \end{aligned}x1=rx2=θx3=ϕ
In n n nnn-dimensional space we can write x i x i x^(i)x^{i}xi, where i i iii represents any of the coordinate indices 1 , 2 , 3 , … , n 1 , 2 , 3 , … , n 1,2,3,dots,n1,2,3, \ldots, n1,2,3,…,n.

1.2 Vectors

Figure 1.1: Vector field v ( x , y ) = y e ^ x − x e ^ y v ( x , y ) = y e ^ x − x e ^ y v(x,y)=y hat(e)_(x)-x hat(e)_(y)\mathbf{v}(x, y)=y \hat{\mathbf{e}}_{x}-x \hat{\mathbf{e}}_{y}v(x,y)=ye^x−xe^y.
Vectors will usually be denoted by lower case upright boldface type ( u u u\mathbf{u}u, v v v\mathbf{v}v, etc). Sometimes we'll number them ( v 1 , v 2 v 1 , v 2 v_(1),v_(2)\mathbf{v}_{1}, \mathbf{v}_{2}v1,v2, etc).
If a vector v v v\mathbf{v}v has components v 1 , v 2 , v 3 v 1 , v 2 , v 3 v^(1),v^(2),v^(3)v^{1}, v^{2}, v^{3}v1,v2,v3, we'll write v = ( v 1 , v 2 , v 3 ) v = v 1 , v 2 , v 3 v=(v^(1),v^(2),v^(3))\mathbf{v}=\left(v^{1}, v^{2}, v^{3}\right)v=(v1,v2,v3).
In R n R n R^(n)\mathbb{R}^{n}Rn a vector can be written as a linear combination of the orthogonal standard basis vectors
(1.2.1) e 1 = ( 1 , 0 , … , 0 , 0 ) , e 2 = ( 0 , 1 , … , 0 , 0 ) , … , e n = ( 0 , 0 , … , 0 , 1 ) (1.2.1) e 1 = ( 1 , 0 , … , 0 , 0 ) , e 2 = ( 0 , 1 , … , 0 , 0 ) , … , e n = ( 0 , 0 , … , 0 , 1 ) {:(1.2.1)e_(1)=(1","0","dots","0","0)","e_(2)=(0","1","dots","0","0)","dots","e_(n)=(0","0","dots","0","1):}\begin{equation*} \mathbf{e}_{1}=(1,0, \ldots, 0,0), \mathbf{e}_{2}=(0,1, \ldots, 0,0), \ldots, \mathbf{e}_{n}=(0,0, \ldots, 0,1) \tag{1.2.1} \end{equation*}(1.2.1)e1=(1,0,…,0,0),e2=(0,1,…,0,0),…,en=(0,0,…,0,1)
pointing along the x 1 , x 2 , … , x n x 1 , x 2 , … , x n x^(1),x^(2),dots,x^(n)x^{1}, x^{2}, \ldots, x^{n}x1,x2,…,xn axes. So, for example, a vector in R 4 R 4 R^(4)\mathbb{R}^{4}R4 (four-dimensional Euclidean space) can be written as
v = v 1 e 1 + v 2 e 2 + v 3 e 3 + v 4 e 4 v = v 1 e 1 + v 2 e 2 + v 3 e 3 + v 4 e 4 v=v^(1)e_(1)+v^(2)e_(2)+v^(3)e_(3)+v^(4)e_(4)\mathbf{v}=v^{1} \mathbf{e}_{1}+v^{2} \mathbf{e}_{2}+v^{3} \mathbf{e}_{3}+v^{4} \mathbf{e}_{4}v=v1e1+v2e2+v3e3+v4e4
which, using index notation (where i i iii takes the value of 1 , 2 , 3 1 , 2 , 3 1,2,31,2,31,2,3 or 4 ), can be shortened to v = v i e i . 1 v = v i e i . 1 v=v^(i)e_(i).^(1)\mathbf{v}=v^{i} \mathrm{e}_{i} .^{1}v=viei.1
When using a Cartesian coordinate system, we'll use the conventional notation e ^ x , e ^ y , e ^ z e ^ x , e ^ y , e ^ z hat(e)_(x), hat(e)_(y), hat(e)_(z)\hat{\mathbf{e}}_{x}, \hat{\mathbf{e}}_{y}, \hat{\mathbf{e}}_{z}e^x,e^y,e^z for the standard unit basis vectors pointing along the x , y , z x , y , z x,y,zx, y, zx,y,z coordinate axes.
A vector field associates a vector to each point in a space. A wind map, for example, indicating wind speed and direction for different points on the Earth's surface. Figure 1.1 shows a plot of the R 2 R 2 R^(2)\mathbb{R}^{2}R2 vector field v ( x , y ) = y e ^ x − x e ^ y v ( x , y ) = y e ^ x − x e ^ y v(x,y)=y hat(e)_(x)-x hat(e)_(y)\mathbf{v}(x, y)=y \hat{\mathbf{e}}_{x}-x \hat{\mathbf{e}}_{y}v(x,y)=ye^x−xe^y. The general form of a three-dimensional vector field in Cartesian coordinates is
(1.2.2) v ( x , y , z ) = f 1 ( x , y , z ) e ^ x + f 2 ( x , y , z ) e ^ y + f 3 ( x , y , z ) e ^ z (1.2.2) v ( x , y , z ) = f 1 ( x , y , z ) e ^ x + f 2 ( x , y , z ) e ^ y + f 3 ( x , y , z ) e ^ z {:(1.2.2)v(x","y","z)=f_(1)(x","y","z) hat(e)_(x)+f_(2)(x","y","z) hat(e)_(y)+f_(3)(x","y","z) hat(e)_(z):}\begin{equation*} \mathbf{v}(x, y, z)=f_{1}(x, y, z) \hat{\mathbf{e}}_{x}+f_{2}(x, y, z) \hat{\mathbf{e}}_{y}+f_{3}(x, y, z) \hat{\mathbf{e}}_{z} \tag{1.2.2} \end{equation*}(1.2.2)v(x,y,z)=f1(x,y,z)e^x+f2(x,y,z)e^y+f3(x,y,z)e^z
where the components of the vector at any point are given by the smooth (ie differentiable) scalar functions f 1 ( x , y , z ) , f 2 ( x , y , z ) f 1 ( x , y , z ) , f 2 ( x , y , z ) f_(1)(x,y,z),f_(2)(x,y,z)f_{1}(x, y, z), f_{2}(x, y, z)f1(x,y,z),f2(x,y,z) and f 3 ( x , y , z ) f 3 ( x , y , z ) f_(3)(x,y,z)f_{3}(x, y, z)f3(x,y,z). Or, more simply, we can write ( 1.2 .2 ) ( 1.2 .2 ) (1.2.2)(1.2 .2)(1.2.2) as
v ( x , y , z ) = f 1 e ^ x + f 2 e ^ y + f 3 e ^ z v ( x , y , z ) = f 1 e ^ x + f 2 e ^ y + f 3 e ^ z v(x,y,z)=f_(1) hat(e)_(x)+f_(2) hat(e)_(y)+f_(3) hat(e)_(z)\mathbf{v}(x, y, z)=f_{1} \hat{\mathbf{e}}_{x}+f_{2} \hat{\mathbf{e}}_{y}+f_{3} \hat{\mathbf{e}}_{z}v(x,y,z)=f1e^x+f2e^y+f3e^z

1.3 Tangent vectors

Figure 1.2: Tangent space to sphere and curve.
A vector field on a manifold is constructed by assigning a tangent vector to every point. Intuitively, we can regard a tangent vector as a vector that, at a given point, is tangent to a curve, surface or higher dimensional space. The set of tangent vectors at a point P P PPP is a vector space 2 2 ^(2){ }^{2}2 that is called the tangent space T P M T P M T_(P)MT_{P} MTPM of the manifold M M MMM at P . T P M P . T P M P.T_(P)MP . T_{P} MP.TPM has the same dimension as M M MMM. A tangent space T P M T P M T_(P)MT_{P} MTPM is attached to every point on M M MMM.
In Figure 1.2, the tangent vectors at point P P PPP on the surface of a sphere S S SSS (embedded in R 3 R 3 R^(3)\mathbb{R}^{3}R3 ) lie in a plane - the tangent space T P S T P S T_(P)ST_{P} STPS - that is tangent to the sphere at P P PPP. Also shown is the curve C C CCC (embedded in R 2 R 2 R^(2)\mathbb{R}^{2}R2 ) of y = f ( x ) = x 2 y = f ( x ) = x 2 y=f(x)=x^(2)y=f(x)=x^{2}y=f(x)=x2. The line tangent to the curve at point P P PPP is the tangent space T P C T P C T_(P)CT_{P} CTPC to the curve at P P PPP, ie T P C T P C T_(P)CT_{P} CTPC is home to all the tangent vectors at P P PPP. One way to parameterise C C CCC would be with x = t x = t x=tx=tx=t and y = t 2 y = t 2 y=t^(2)y=t^{2}y=t2. If point Q ( x , y ) = ( 2 , 4 ) Q ( x , y ) = ( 2 , 4 ) Q(x,y)=(2,4)Q(x, y)=(2,4)Q(x,y)=(2,4), the tangent vector v ( t ) v ( t ) v(t)\mathbf{v}(t)v(t) at T ( 2 , 4 ) C T ( 2 , 4 ) C T_((2,4))CT_{(2,4)} CT(2,4)C is then
v = d x d t e ^ x + d y d t e ^ y = e ^ x + 2 t e ^ y = e ^ x + 4 e ^ y . v = d x d t e ^ x + d y d t e ^ y = e ^ x + 2 t e ^ y = e ^ x + 4 e ^ y . {:[v=(dx)/(dt) hat(e)_(x)+(dy)/(dt) hat(e)_(y)= hat(e)_(x)+2t hat(e)_(y)],[= hat(e)_(x)+4 hat(e)_(y).]:}\begin{gathered} \mathbf{v}=\frac{d x}{d t} \hat{\mathbf{e}}_{x}+\frac{d y}{d t} \hat{\mathbf{e}}_{y}=\hat{\mathbf{e}}_{x}+2 t \hat{\mathbf{e}}_{y} \\ =\hat{\mathbf{e}}_{x}+4 \hat{\mathbf{e}}_{y} . \end{gathered}v=dxdte^x+dydte^y=e^x+2te^y=e^x+4e^y.
Or, in component form, v = ( 1 , 4 ) v = ( 1 , 4 ) v=(1,4)\mathbf{v}=(1,4)v=(1,4). Different parameterisations of C C CCC would give different tangent vectors. For example, we could use x = 3 t x = 3 t x=3tx=3 tx=3t and y = 9 t 2 y = 9 t 2 y=9t^(2)y=9 t^{2}y=9t2, or x = 10 t x = 10 t x=10 tx=10 tx=10t and y = 100 t 2 y = 100 t 2 y=100t^(2)y=100 t^{2}y=100t2. However, at point ( 2 , 4 ) ( 2 , 4 ) (2,4)(2,4)(2,4) the resulting tangent vectors would all be multiples of v = e ^ x + 4 e ^ y v = e ^ x + 4 e ^ y v= hat(e)_(x)+4 hat(e)_(y)\mathbf{v}=\hat{\mathbf{e}}_{x}+4 \hat{\mathbf{e}}_{y}v=e^x+4e^y.
The sphere S S SSS and curve C C CCC shown in Figure 1.2 are both embedded in Euclidean space, as are their respective tangent spaces T P S T P S T_(P)ST_{P} STPS (a plane) and T P C T P C T_(P)CT_{P} CTPC (a straight line). This notion of embedded manifolds and tangent spaces can be extended to higher dimensions. For example, a three-dimensional manifold M M MMM embedded in R 4 R 4 R^(4)\mathbb{R}^{4}R4 would, at a point P P PPP, have a tangent space - a hyperplane - also embedded in R 4 R 4 R^(4)\mathbb{R}^{4}R4 and consisting of all the vectors tangent to M M MMM at P P PPP.
For a more general manifold M M MMM that isn't embedded in some R n R n R^(n)\mathbb{R}^{n}Rn, a tangent vector at point P P PPP may still be regarded as a vector tangent to a smooth curve that passes through P P PPP. However, these 'intrinsic' tangent vectors (think of them as infinitesimally tiny arrows with their base points attached to P P PPP ) are defined to live on the manifold M M MMM itself, not embedded in some higher dimensional space. The tangent space T P M T P M T_(P)MT_{P} MTPM then consists of the set of vectors at P P PPP tangent to all smooth curves passing through P P PPP.
  • Differential forms and tangent vectors live on manifolds. Differential forms and tangent vectors act on each other (at a point, to give a number). Therefore, in this book most of the vectors we discuss are tangent vectors, even if we don't explicitly refer to them as such. Mathematicians have devised various precise but equivalent definitions of tangent vectors on manifolds that, at our level, we don't need to worry about. The essential point is that differential forms act on tangent vectors.

1.4 Parameterisation

It is often convenient to describe manifolds parametrically. A curve can be parameterised using a single parameter. For example, a helix in R 3 R 3 R^(3)\mathbb{R}^{3}R3 has the parametric
independent vectors that spans V V VVV, meaning any vector in V V VVV can be written in terms of the basis vectors. The number of vectors in a basis for V V VVV is called the dimension of V V VVV. Vector spaces are ubiquitous in mathematics and physics.
equations (with parameter t t ttt ) x = cos ⁡ t , y = sin ⁡ t , z = t x = cos ⁡ t , y = sin ⁡ t , z = t x=cos t,y=sin t,z=tx=\cos t, y=\sin t, z=tx=cos⁡t,y=sin⁡t,z=t, which we can also write as a vector equation, using the Greek letter Phi ( Φ ) Phi ( Φ ) Phi(Phi)\operatorname{Phi}(\boldsymbol{\Phi})Phi(Φ), as
Φ ( t ) = cos ⁡ t e ^ x + sin ⁡ t e ^ y + t e ^ z Φ ( t ) = cos ⁡ t e ^ x + sin ⁡ t e ^ y + t e ^ z Phi(t)=cos t hat(e)_(x)+sin t hat(e)_(y)+t hat(e)_(z)\boldsymbol{\Phi}(t)=\cos t \hat{\mathbf{e}}_{x}+\sin t \hat{\mathbf{e}}_{y}+t \hat{\mathbf{e}}_{z}Φ(t)=cos⁡te^x+sin⁡te^y+te^z
Or, in component form:
Φ ( t ) = ( cos ⁡ t , sin ⁡ t , t ) Φ ( t ) = ( cos ⁡ t , sin ⁡ t , t ) Phi(t)=(cos t,sin t,t)\mathbf{\Phi}(t)=(\cos t, \sin t, t)Φ(t)=(cos⁡t,sin⁡t,t)
A surface can be parameterised using two parameters. The top half of a unit radius sphere in R 3 R 3 R^(3)\mathbb{R}^{3}R3, for example, can be parameterised (with parameters u u uuu and v v vvv ) by x = v cos ⁡ u , y = v sin ⁡ u , z = 1 − v 2 x = v cos ⁡ u , y = v sin ⁡ u , z = 1 − v 2 x=v cos u,y=v sin u,z=sqrt(1-v^(2))x=v \cos u, y=v \sin u, z=\sqrt{1-v^{2}}x=vcos⁡u,y=vsin⁡u,z=1−v2 for 0 ≤ u ≤ 2 π , 0 ≤ v ≤ 1 0 ≤ u ≤ 2 π , 0 ≤ v ≤ 1 0 <= u <= 2pi,0 <= v <= 10 \leq u \leq 2 \pi, 0 \leq v \leq 10≤u≤2π,0≤v≤1. We can write this as the vector equation
Φ ( u , v ) = v cos ⁡ u e ^ x + v sin ⁡ u e ^ y + 1 − v 2 e ^ z Φ ( u , v ) = v cos ⁡ u e ^ x + v sin ⁡ u e ^ y + 1 − v 2 e ^ z Phi(u,v)=v cos u hat(e)_(x)+v sin u hat(e)_(y)+sqrt(1-v^(2)) hat(e)_(z)\boldsymbol{\Phi}(u, v)=v \cos u \hat{\mathbf{e}}_{x}+v \sin u \hat{\mathbf{e}}_{y}+\sqrt{1-v^{2}} \hat{\mathbf{e}}_{z}Φ(u,v)=vcos⁡ue^x+vsin⁡ue^y+1−v2e^z
Or, in component form:
Φ ( u , v ) = ( v cos ⁡ u , v sin ⁡ u , 1 − v 2 ) Φ ( u , v ) = v cos ⁡ u , v sin ⁡ u , 1 − v 2 Phi(u,v)=(v cos u,v sin u,sqrt(1-v^(2)))\Phi(u, v)=\left(v \cos u, v \sin u, \sqrt{1-v^{2}}\right)Φ(u,v)=(vcos⁡u,vsin⁡u,1−v2)

1.5 Notation for forms

Differential forms will usually be denoted by lower case Greek letters ( ω , ν ω , ν omega,nu\omega, \nuω,ν, etc).
Sometimes we'll number them ( ω 1 , ω 2 , ω 3 ω 1 , ω 2 , ω 3 (omega_(1),omega_(2),omega_(3):}\left(\omega_{1}, \omega_{2}, \omega_{3}\right.(ω1,ω2,ω3, etc) ) ) ))).
We'll denote a 1 -form ω ω omega\omegaω acting on a vector v v v\mathbf{v}v as ω ( v ) ω ( v ) omega(v)\omega(\mathbf{v})ω(v). A 2 -form ω ω omega\omegaω acting on two vectors u u u\mathbf{u}u and v v v\mathbf{v}v will be shown as ω ( u , v ) ω ( u , v ) omega(u,v)\omega(\mathbf{u}, \mathbf{v})ω(u,v), and so on.

1.6 Cross product

Figure 1.3: Cross product of vectors u u u\mathbf{u}u and v v v\mathbf{v}v.
The cross product u × v u × v uxxv\mathbf{u} \times \mathbf{v}u×v of two three-dimensional vectors u u u\mathbf{u}u and v v v\mathbf{v}v is itself a vector (actually, a pseudovector, but we don't need to worry about that) and is perpendicular to u u u\mathbf{u}u and v v v\mathbf{v}v, as shown in Figure 1.3. By convention the direction of u × v u × v uxxv\mathbf{u} \times \mathbf{v}u×v is given by the right-hand rule. Hold the thumb, index finger and second finger of your right hand perpendicular to each other. If your index finger points in the direction of u u u\mathbf{u}u and your second finger points in the direction of v v v\mathbf{v}v, then your thumb points in the direction of u × v u × v uxxv\mathbf{u} \times \mathbf{v}u×v. The cross product v × u = − u × v v × u = − u × v vxxu=-uxxv\mathbf{v} \times \mathbf{u}=-\mathbf{u} \times \mathbf{v}v×u=−u×v is in the opposite direction to u × v u × v uxxv\mathbf{u} \times \mathbf{v}u×v.
If u = u 1 e ^ x + u 2 e ^ y + u 3 e ^ z u = u 1 e ^ x + u 2 e ^ y + u 3 e ^ z u=u^(1) hat(e)_(x)+u^(2) hat(e)_(y)+u^(3) hat(e)_(z)\mathbf{u}=u^{1} \hat{\mathbf{e}}_{x}+u^{2} \hat{\mathbf{e}}_{y}+u^{3} \hat{\mathbf{e}}_{z}u=u1e^x+u2e^y+u3e^z and v = v 1 e ^ x + v 2 e ^ y + v 3 e ^ z v = v 1 e ^ x + v 2 e ^ y + v 3 e ^ z v=v^(1) hat(e)_(x)+v^(2) hat(e)_(y)+v^(3) hat(e)_(z)\mathbf{v}=v^{1} \hat{\mathbf{e}}_{x}+v^{2} \hat{\mathbf{e}}_{y}+v^{3} \hat{\mathbf{e}}_{z}v=v1e^x+v2e^y+v3e^z, the cross product of u u u\mathbf{u}u and v v v\mathbf{v}v is given by
(1.6.1) u × v = ( u 2 v 3 − u 3 v 2 ) e ^ x + ( u 3 v 1 − u 1 v 3 ) e ^ y + ( u 1 v 2 − u 2 v 1 ) e ^ z (1.6.1) u × v = u 2 v 3 − u 3 v 2 e ^ x + u 3 v 1 − u 1 v 3 e ^ y + u 1 v 2 − u 2 v 1 e ^ z {:(1.6.1)uxxv=(u^(2)v^(3)-u^(3)v^(2)) hat(e)_(x)+(u^(3)v^(1)-u^(1)v^(3)) hat(e)_(y)+(u^(1)v^(2)-u^(2)v^(1)) hat(e)_(z):}\begin{equation*} \mathbf{u} \times \mathbf{v}=\left(u^{2} v^{3}-u^{3} v^{2}\right) \hat{\mathbf{e}}_{x}+\left(u^{3} v^{1}-u^{1} v^{3}\right) \hat{\mathbf{e}}_{y}+\left(u^{1} v^{2}-u^{2} v^{1}\right) \hat{\mathbf{e}}_{z} \tag{1.6.1} \end{equation*}(1.6.1)u×v=(u2v3−u3v2)e^x+(u3v1−u1v3)e^y+(u1v2−u2v1)e^z
Or, as a 3 × 3 3 × 3 3xx33 \times 33×3 determinant (see section 1.7 ),
(1.6.2) u × v = | i j k u 1 u 2 u 3 v 1 v 2 v 3 | (1.6.2) u × v = i j k u 1 u 2 u 3 v 1 v 2 v 3 {:(1.6.2)uxxv=|[i,j,k],[u^(1),u^(2),u^(3)],[v^(1),v^(2),v^(3)]|:}\mathbf{u} \times \mathbf{v}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \tag{1.6.2}\\ u^{1} & u^{2} & u^{3} \\ v^{1} & v^{2} & v^{3} \end{array}\right|(1.6.2)u×v=|ijku1u2u3v1v2v3|
The area of the parallelogram spanned by u u u\mathbf{u}u and v v v\mathbf{v}v is given by the magnitude of the cross product ‖ u × v ‖ ‖ u × v ‖ ||uxxv||\|\mathbf{u} \times \mathbf{v}\|‖u×v‖.

1.7 Determinants

The determinant is a single number that can be calculated from a square matrix. The determinant has all sorts of weird and wonderful properties, but its importance as far as we're concerned is that it allows us to carry out calculations involving the wedge product. As we'll see, determinants are alternating multilinear functions of their columns (or rows). Being alternating and multilinear are the two key properties of determinants that transfer over to differential forms.
The determinant of a matrix A A AAA is denoted by det ( A ) det ( A ) det(A)\operatorname{det}(A)det(A) or | A | | A | |A||A||A|.

1.7.1 The determinant of a 1 × 1 1 × 1 1xx11 \times 11×1 matrix

For a matrix
A = [ a ] A = [ a ] A=[a]A=[a]A=[a]
the determinant is simply equal to a a aaa, ie
| A | = a | A | = a |A|=a|A|=a|A|=a
Figure 1.4: Parallelogram spanned by u u u\mathbf{u}u and v v v\mathbf{v}v.

1.7.2 The determinant of a 2 × 2 2 × 2 2xx22 \times 22×2 matrix

For a matrix
A = [ a b c d ] A = a      b c      d A=[[a,b],[c,d]]A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]A=[abcd]
the determinant is given by
| A | = a d − c b | A | = a d − c b |A|=ad-cb|A|=a d-c b|A|=ad−cb
Figure 1.4 shows a parallelogram spanned by two vectors: u = u 1 e ^ x + u 2 e ^ y u = u 1 e ^ x + u 2 e ^ y u=u^(1) hat(e)_(x)+u^(2) hat(e)_(y)\mathbf{u}=u^{1} \hat{\mathbf{e}}_{x}+u^{2} \hat{\mathbf{e}}_{y}u=u1e^x+u2e^y and v = v 1 e ^ x + v 2 e ^ y v = v 1 e ^ x + v 2 e ^ y v=v^(1) hat(e)_(x)+v^(2) hat(e)_(y)\mathbf{v}=v^{1} \hat{\mathbf{e}}_{x}+v^{2} \hat{\mathbf{e}}_{y}v=v1e^x+v2e^y. If we write u u u\mathbf{u}u and v v v\mathbf{v}v as column vectors, we get the matrix
[ u 1 v 1 u 2 v 2 ] u 1 v 1 u 2 v 2 [[u^(1),v^(1)],[u^(2),v^(2)]]\left[\begin{array}{cc} u^{1} & v^{1} \\ u^{2} & v^{2} \end{array}\right][u1v1u2v2]
The determinant of this matrix gives the area of the spanned parallelogram:
| u 1 v 1 u 2 v 2 | = u 1 v 2 − u 2 v 1 u 1 v 1 u 2 v 2 = u 1 v 2 − u 2 v 1 |[u^(1),v^(1)],[u^(2),v^(2)]|=u^(1)v^(2)-u^(2)v^(1)\left|\begin{array}{cc} u^{1} & v^{1} \\ u^{2} & v^{2} \end{array}\right|=u^{1} v^{2}-u^{2} v^{1}|u1v1u2v2|=u1v2−u2v1
Determinants are alternating, which means if we interchange any two columns (or rows), the determinant changes sign. So if we swap the vectors, we swap the sign of the area, ie
| v 1 u 1 v 2 u 2 | = v 1 u 2 − v 2 u 1 v 1      u 1 v 2      u 2 = v 1 u 2 − v 2 u 1 |[v^(1),u^(1)],[v^(2),u^(2)]|=v^(1)u^(2)-v^(2)u^(1)\left|\begin{array}{ll} v^{1} & u^{1} \\ v^{2} & u^{2} \end{array}\right|=v^{1} u^{2}-v^{2} u^{1}|v1u1v2u2|=v1u2−v2u1
So we say the determinant gives the signed area of the spanned parallelogram. We'll revisit this notion of things being alternating - the sign changing when the order of the vectors is flipped - when we see how 2 -forms and higher degree forms act on vectors to spit out a number. The number will be positive or negative depending on the order that we feed the vectors to the differential form.
Determinants are also multilinear in terms of their columns (or rows). This means if we multiply the vector u = ( u 1 , u 2 ) u = u 1 , u 2 u=(u^(1),u^(2))\mathbf{u}=\left(u^{1}, u^{2}\right)u=(u1,u2) by a constant k k kkk, the resulting determinant
| k u 1 v 1 k u 2 v 2 | = k ( u 1 v 2 − u 2 v 1 ) k u 1      v 1 k u 2      v 2 = k u 1 v 2 − u 2 v 1 |[ku^(1),v^(1)],[ku^(2),v^(2)]|=k(u^(1)v^(2)-u^(2)v^(1))\left|\begin{array}{ll} k u^{1} & v^{1} \\ k u^{2} & v^{2} \end{array}\right|=k\left(u^{1} v^{2}-u^{2} v^{1}\right)|ku1v1ku2v2|=k(u1v2−u2v1)
is k k kkk times the size of the original determinant u 1 v 2 − u 2 v 1 u 1 v 2 − u 2 v 1 u^(1)v^(2)-u^(2)v^(1)u^{1} v^{2}-u^{2} v^{1}u1v2−u2v1.

1.7.3 The determinant of a 3 × 3 3 × 3 3xx33 \times 33×3 matrix

For a matrix
A = [ a b c d e f g h i ] A = a      b      c d      e      f g      h      i A=[[a,b,c],[d,e,f],[g,h,i]]A=\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right]A=[abcdefghi]
the determinant is given by
| A | = a ( e i − f h ) − b ( d i − f g ) + c ( d h − e g ) | A | = a ( e i − f h ) − b ( d i − f g ) + c ( d h − e g ) |A|=a(ei-fh)-b(di-fg)+c(dh-eg)|A|=a(e i-f h)-b(d i-f g)+c(d h-e g)|A|=a(ei−fh)−b(di−fg)+c(dh−eg)
which is equivalent to
| A | = a | e f h i | − b | d f g i | + c | d e g h | . | A | = a e f h i − b d f g i + c d e g h . |A|=a|[e,f],[h,i]|-b|[d,f],[g,i]|+c|[d,e],[g,h]|.|A|=a\left|\begin{array}{cc} e & f \\ h & i \end{array}\right|-b\left|\begin{array}{cc} d & f \\ g & i \end{array}\right|+c\left|\begin{array}{cc} d & e \\ g & h \end{array}\right| .|A|=a|efhi|−b|dfgi|+c|degh|.
There are alternative forms, including
(1.7.1) | A | = a ( e i − f h ) + b ( f g − d i ) + c ( d h − e g ) . (1.7.1) | A | = a ( e i − f h ) + b ( f g − d i ) + c ( d h − e g ) . {:(1.7.1)|A|=a(ei-fh)+b(fg-di)+c(dh-eg).:}\begin{equation*} |A|=a(e i-f h)+b(f g-d i)+c(d h-e g) . \tag{1.7.1} \end{equation*}(1.7.1)|A|=a(ei−fh)+b(fg−di)+c(dh−eg).
The volume of a parallelepiped spanned by three vectors, u = u 1 e ^ x + u 2 e ^ y + u 3 e ^ z u = u 1 e ^ x + u 2 e ^ y + u 3 e ^ z u=u^(1) hat(e)_(x)+u^(2) hat(e)_(y)+u^(3) hat(e)_(z)\mathbf{u}=u^{1} \hat{\mathbf{e}}_{x}+u^{2} \hat{\mathbf{e}}_{y}+u^{3} \hat{\mathbf{e}}_{z}u=u1e^x+u2e^y+u3e^z, v = v 1 e ^ x + v 2 e ^ y + v 3 e ^ z v = v 1 e ^ x + v 2 e ^ y + v 3 e ^ z v=v^(1) hat(e)_(x)+v^(2) hat(e)_(y)+v^(3) hat(e)_(z)\mathbf{v}=v^{1} \hat{\mathbf{e}}_{x}+v^{2} \hat{\mathbf{e}}_{y}+v^{3} \hat{\mathbf{e}}_{z}v=v1e^x+v2e^y+v3e^z and w = w 1 e ^ x + w 2 e ^ y + w 3 e ^ z w = w 1 e ^ x + w 2 e ^ y + w 3 e ^ z w=w^(1) hat(e)_(x)+w^(2) hat(e)_(y)+w^(3) hat(e)_(z)\mathbf{w}=w^{1} \hat{\mathbf{e}}_{x}+w^{2} \hat{\mathbf{e}}_{y}+w^{3} \hat{\mathbf{e}}_{z}w=w1e^x+w2e^y+w3e^z, as shown in Figure 1.5 , is the determinant of the column vector matrix
[ u 1 v 1 w 1 u 2 v 2 w 2 u 3 v 3 w 3 ] u 1 v 1 w 1 u 2 v 2 w 2 u 3 v 3 w 3 [[u^(1),v^(1),w^(1)],[u^(2),v^(2),w^(2)],[u^(3),v^(3),w^(3)]]\left[\begin{array}{ccc} u^{1} & v^{1} & w^{1} \\ u^{2} & v^{2} & w^{2} \\ u^{3} & v^{3} & w^{3} \end{array}\right][u1v1w1u2v2w2u3v3w3]
This is a signed volume because if we swap any two column vectors, the sign of the determinate changes.
Figure 1.5: Parallelepiped spanned by u , v u , v u,v\mathbf{u}, \mathbf{v}u,v and w w w\mathbf{w}w.

1.7.4 The determinate of an n × n n × n n xx nn \times nn×n matrix, where n > 3 n > 3 n > 3n>3n>3

Tricky to visualise, trickier to draw, a parallelotope is a generalization of the twodimensional parallelogram and three-dimensional parallelepiped into higher dimension. The determinate of an n × n n × n n xx nn \times nn×n matrix, where n > 3 n > 3 n > 3n>3n>3, gives the signed volume of the n n nnn-dimensional parallelotope spanned by the matrix's n n nnn column vectors.
Example 1.1. In R 4 R 4 R^(4)\mathbb{R}^{4}R4, find the volume of the parallelotope spanned by the vectors v 1 = ( 3 , 0 , 0 , 0 ) , v 2 = ( 0 , 3 , 0 , 0 ) , v 3 = ( 0 , 0 , 3 , 0 ) v 1 = ( 3 , 0 , 0 , 0 ) , v 2 = ( 0 , 3 , 0 , 0 ) , v 3 = ( 0 , 0 , 3 , 0 ) v_(1)=(3,0,0,0),v_(2)=(0,3,0,0),v_(3)=(0,0,3,0)\mathbf{v}_{1}=(3,0,0,0), \mathbf{v}_{2}=(0,3,0,0), \mathbf{v}_{3}=(0,0,3,0)v1=(3,0,0,0),v2=(0,3,0,0),v3=(0,0,3,0) and v 4 = ( 0 , 0 , 0 , 3 ) v 4 = ( 0 , 0 , 0 , 3 ) v_(4)=(0,0,0,3)\mathbf{v}_{4}=(0,0,0,3)v4=(0,0,0,3).
This parallelotope is actually a four-dimensional cube of side length 3 , so the volume equals 3 4 = 81 3 4 = 81 3^(4)=813^{4}=8134=81. The volume is also given by the determinant of the column vector matrix
[ 3 0 0 0 0 3 0 0 0 0 3 0 0 0 0 3 ] = 81 3      0      0      0 0      3      0      0 0      0      3      0 0      0      0      3 = 81 [[3,0,0,0],[0,3,0,0],[0,0,3,0],[0,0,0,3]]=81\left[\begin{array}{llll} 3 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{array}\right]=81[3000030000300003]=81
You can check this result using the WolframAlpha online calculator [22] by typing the following into the input box and hitting Enter:
determinant {{3,0,0,0},{0,3,0,0},{0,0,3,0},{0,0,0,3}}
It's easy enough to check this is a signed volume by swapping any two column vectors and seeing that the sign of the determinate changes.

2 What are differential forms?

Differential forms are linear or, more generally, multilinear alternating functions of tangent vectors.
  • A 0 -form is a special case and is simply a smooth function. A 0-forms eats a point and returns a number.
  • A 1-form eats a vector and returns a number.
  • A 2-form eats two vectors and returns a number.
  • A 3 -form eats three vectors and returns a number.
  • A k k kkk-form eats k k kkk vectors and returns a number.
In terms of identification, differential forms can be generally recognised as things containing differentials such as d x , d y d x , d y dx,dyd x, d ydx,dy and d z d z dzd zdz. So these are 1-forms (each term contains one differential)
ω = 2 x d x + 3 y d y − d z ω = x d y ω = d x ω = 2 x d x + 3 y d y − d z ω = x d y ω = d x {:[omega=2xdx+3ydy-dz],[omega=xdy],[omega=dx]:}\begin{gathered} \omega=2 x d x+3 y d y-d z \\ \omega=x d y \\ \omega=d x \end{gathered}ω=2xdx+3ydy−dzω=xdyω=dx
This is a 2 -form
ω = 8 d y ∧ d z ω = 8 d y ∧ d z omega=8dy^^dz\omega=8 d y \wedge d zω=8dy∧dz
and this is a 3 -form
ω = − 7 z d x ∧ d y ∧ d z ω = − 7 z d x ∧ d y ∧ d z omega=-7zdx^^dy^^dz\omega=-7 z d x \wedge d y \wedge d zω=−7zdx∧dy∧dz
where the ∧ ∧ ^^\wedge∧ symbol denotes a type of multiplication called the wedge product (details to follow).
A vector field on a manifold gives a vector (actually, a tangent vector) at every point. So, for example, say a point P P PPP on M M MMM, a region of R 3 R 3 R^(3)\mathbb{R}^{3}R3, has coordinates x = 3 , y = 5 , z = 1 x = 3 , y = 5 , z = 1 x=3,y=5,z=1x=3, y=5, z=1x=3,y=5,z=1. At P P PPP the vector field
v = 7 x e ^ x + 4 y e ^ y + 2 e ^ z v = 7 x e ^ x + 4 y e ^ y + 2 e ^ z v=7x hat(e)_(x)+4y hat(e)_(y)+2 hat(e)_(z)\mathbf{v}=7 x \hat{\mathbf{e}}_{x}+4 y \hat{\mathbf{e}}_{y}+2 \hat{\mathbf{e}}_{z}v=7xe^x+4ye^y+2e^z
gives the (tangent) vector
v = ( 7 × 3 ) e ^ x + ( 4 × 5 ) e ^ y + 2 e ^ z v = 21 e ^ x + 20 e ^ y + 2 e ^ z v = ( 7 × 3 ) e ^ x + ( 4 × 5 ) e ^ y + 2 e ^ z v = 21 e ^ x + 20 e ^ y + 2 e ^ z {:[v=(7xx3) hat(e)_(x)+(4xx5) hat(e)_(y)+2 hat(e)_(z)],[v=21 hat(e)_(x)+20 hat(e)_(y)+2 hat(e)_(z)]:}\begin{gathered} \mathbf{v}=(7 \times 3) \hat{\mathbf{e}}_{x}+(4 \times 5) \hat{\mathbf{e}}_{y}+2 \hat{\mathbf{e}}_{z} \\ \mathbf{v}=21 \hat{\mathbf{e}}_{x}+20 \hat{\mathbf{e}}_{y}+2 \hat{\mathbf{e}}_{z} \end{gathered}v=(7×3)e^x+(4×5)e^y+2e^zv=21e^x+20e^y+2e^z
Similarly, a differential form on a manifold can be thought of as a differential form field, that gives a particular differential form at every point. For example, at the same point P ( 3 , 5 , 1 ) P ( 3 , 5 , 1 ) P(3,5,1)P(3,5,1)P(3,5,1) on M M MMM the above 1-form field, ω = 2 x d x + 3 y d y − d z ω = 2 x d x + 3 y d y − d z omega=2xdx+3ydy-dz\omega=2 x d x+3 y d y-d zω=2xdx+3ydy−dz, gives the particular 1 -form
ω = ( 2 × 3 ) d x + ( 3 × 5 ) d y − d z ω = 6 d x + 15 d y − d z ω = ( 2 × 3 ) d x + ( 3 × 5 ) d y − d z ω = 6 d x + 15 d y − d z {:[omega=(2xx3)dx+(3xx5)dy-dz],[omega=6dx+15 dy-dz]:}\begin{gathered} \omega=(2 \times 3) d x+(3 \times 5) d y-d z \\ \omega=6 d x+15 d y-d z \end{gathered}ω=(2×3)dx+(3×5)dy−dzω=6dx+15dy−dz
As we'll see, if the tangent vector and 1-form are associated with the same point, they can act on each other to give a number.
Following widely accepted practice, we won't use the term 'field', as in 'differential form field'. Instead, we'll let 'differential form' refer to both a differential form at a point and a differential form field on the manifold.
(A hint of what is to come: just as the vectors e ^ x , e ^ y e ^ x , e ^ y hat(e)_(x), hat(e)_(y)\hat{\mathbf{e}}_{x}, \hat{\mathbf{e}}_{y}e^x,e^y and e ^ z e ^ z hat(e)_(z)\hat{\mathbf{e}}_{z}e^z form a set of basis vectors for R 3 R 3 R^(3)\mathbb{R}^{3}R3, the differentials d x , d y d x , d y dx,dyd x, d ydx,dy and d z d z dzd zdz form a set of basis 1 -forms for R 3 R 3 R^(3)\mathbb{R}^{3}R3.)

2.10 -forms

A 0 -form is simply a smooth scalar function. For example:
f ( x , y , z ) = x 2 − z y f ( x , y , z ) = x 2 − z y f(x,y,z)=x^(2)-zyf(x, y, z)=x^{2}-z yf(x,y,z)=x2−zy
which takes a point on a manifold, in this case R 3 R 3 R^(3)\mathbb{R}^{3}R3, and returns a number.

2.2 1-forms

At a point, a 1 -form takes a tangent vector and returns a number. In maths-speak, a 1 -form is a function that requires one vector argument. Not only that, a 1-form ω ω omega\omegaω is a linear real-valued function of vectors, in the sense of
ω ( a u + b v ) = a ω ( u ) + b ω ( v ) ω ( a u + b v ) = a ω ( u ) + b ω ( v ) omega(au+bv)=a omega(u)+b omega(v)\omega(a \mathbf{u}+b \mathbf{v})=a \omega(\mathbf{u})+b \omega(\mathbf{v})ω(au+bv)=aω(u)+bω(v)
where a a aaa and b b bbb are real numbers.
On R 3 R 3 R^(3)\mathbb{R}^{3}R3 a 1-form will include one or more of the differentials d x , d y , d z d x , d y , d z dx,dy,dzd x, d y, d zdx,dy,dz and look something like
ω = f 1 ( x , y , z ) d x + f 2 ( x , y , z ) d y + f 3 ( x , y , z ) d z ω = f 1 ( x , y , z ) d x + f 2 ( x , y , z ) d y + f 3 ( x , y , z ) d z omega=f_(1)(x,y,z)dx+f_(2)(x,y,z)dy+f_(3)(x,y,z)dz\omega=f_{1}(x, y, z) d x+f_{2}(x, y, z) d y+f_{3}(x, y, z) d zω=f1(x,y,z)dx+f2(x,y,z)dy+f3(x,y,z)dz
where f i ( x , y , z ) f i ( x , y , z ) f_(i)(x,y,z)f_{i}(x, y, z)fi(x,y,z) are smooth scalar functions. Or, more simply, we can write
ω = f 1 d x + f 2 d y + f 3 d z ω = f 1 d x + f 2 d y + f 3 d z omega=f_(1)dx+f_(2)dy+f_(3)dz\omega=f_{1} d x+f_{2} d y+f_{3} d zω=f1dx+f2dy+f3dz
(Later on, we'll explore the reasons for this d x , d y , d z d x , d y , d z dx,dy,dzd x, d y, d zdx,dy,dz notation). Of course, we're not limited to Cartesian coordinates in three dimensions. In general coordinates x 1 , x 2 , x 3 , … , x n x 1 , x 2 , x 3 , … , x n x^(1),x^(2),x^(3),dots,x^(n)x^{1}, x^{2}, x^{3}, \ldots, x^{n}x1,x2,x3,…,xn, a 1 -form would be written as
ω = f 1 d x 1 + f 2 d x 2 , … , + f n d x n ω = f 1 d x 1 + f 2 d x 2 , … , + f n d x n omega=f_(1)dx^(1)+f_(2)dx^(2),dots,+f_(n)dx^(n)\omega=f_{1} d x^{1}+f_{2} d x^{2}, \ldots,+f_{n} d x^{n}ω=f1dx1+f2dx2,…,+fndxn
where f i f i f_(i)f_{i}fi are smooth functions of x 1 , x 2 , x 3 , … , x n x 1 , x 2 , x 3 , … , x n x^(1),x^(2),x^(3),dots,x^(n)x^{1}, x^{2}, x^{3}, \ldots, x^{n}x1,x2,x3,…,xn.
The differential of a smooth function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x, y, z)f(x,y,z) is
(2.2.1) d f = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z (2.2.1) d f = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z {:(2.2.1)df=(del f)/(del x)dx+(del f)/(del y)dy+(del f)/(del z)dz:}\begin{equation*} d f=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y+\frac{\partial f}{\partial z} d z \tag{2.2.1} \end{equation*}(2.2.1)df=∂f∂xdx+∂f∂ydy+∂f∂zdz
and is a 1-form. So the differential of the function
f ( x , y , z ) = x z + 4 y 3 + 3 x y f ( x , y , z ) = x z + 4 y 3 + 3 x y f(x,y,z)=xz+4y^(3)+3xyf(x, y, z)=x z+4 y^{3}+3 x yf(x,y,z)=xz+4y3+3xy
is the 1 -form
d f = ( 3 y + z ) d x + ( 3 x + 12 y 2 ) d y + x d z d f = ( 3 y + z ) d x + 3 x + 12 y 2 d y + x d z df=(3y+z)dx+(3x+12y^(2))dy+xdzd f=(3 y+z) d x+\left(3 x+12 y^{2}\right) d y+x d zdf=(3y+z)dx+(3x+12y2)dy+xdz
Even the most humble little d x d x dxd xdx can be regarded as the differential of a function. Suppose we want to change from using x , y , z x , y , z x,y,zx, y, zx,y,z coordinates to some other u , v , w u , v , w u,v,wu, v, wu,v,w coordinate system. If the old x , y , z x , y , z x,y,zx, y, zx,y,z coordinates are functions of the new u , v , w u , v , w u,v,wu, v, wu,v,w coordinates we can find the differential of x ( u , v , w ) x ( u , v , w ) x(u,v,w)x(u, v, w)x(u,v,w), which is
d x = ∂ x ∂ u d u + ∂ x ∂ v d v + ∂ x ∂ w d w d x = ∂ x ∂ u d u + ∂ x ∂ v d v + ∂ x ∂ w d w dx=(del x)/(del u)du+(del x)/(del v)dv+(del x)/(del w)dwd x=\frac{\partial x}{\partial u} d u+\frac{\partial x}{\partial v} d v+\frac{\partial x}{\partial w} d wdx=∂x∂udu+∂x∂vdv+∂x∂wdw
and so on for d y d y dyd ydy and d z d z dzd zdz.
Not all 1-forms are the differentials of functions. For example, there is no function whose differential is the perfectly acceptable 1 -form
ω = x d x + x d y ω = x d x + x d y omega=xdx+xdy\omega=x d x+x d yω=xdx+xdy
Another way of looking at differential forms is simply as objects that can be integrated, including the differential. So in this integral:
∫ a b 6 x d x ⏟ 1 -form ∫ a b   6 x d x ⏟ 1 -form  int_(a)^(b)ubrace(6xdxubrace)_(1"-form ")\int_{a}^{b} \underbrace{6 x d x}_{1 \text {-form }}∫ab6xdx⏟1-form 
6 x d x 6 x d x 6xdx6 x d x6xdx is a 1 -form.
Because we can combine like differentials, we can add two 1-forms together to get another 1-form. For example, we can add 5 d x 5 d x 5dx5 d x5dx and 3 x d x + 8 d y 3 x d x + 8 d y 3xdx+8dy3 x d x+8 d y3xdx+8dy to get ( 3 x + 5 ) d x + 8 d y ( 3 x + 5 ) d x + 8 d y (3x+5)dx+8dy(3 x+5) d x+8 d y(3x+5)dx+8dy.

2.2.1 The meaning of d x i d x i dx^(i)d x^{i}dxi

When we first learned calculus we were taught that d f d f dfd fdf in the aforementioned (2.2.1) differential of a function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x, y, z)f(x,y,z),
d f = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z d f = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z df=(del f)/(del x)dx+(del f)/(del y)dy+(del f)/(del z)dzd f=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y+\frac{\partial f}{\partial z} d zdf=∂f∂xdx+∂f∂ydy+∂f∂zdz
represented an infinitesimal change in f f fff associated with infinitesimal changes d x , d y , d z d x , d y , d z dx,dy,dzd x, d y, d zdx,dy,dz (collectively referred to, using index notation, as d x i d x i dx^(i)d x^{i}dxi ) of the x , y , z x , y , z x,y,zx, y, zx,y,z coordinates.
However, at the start of this chapter we referred to the 1 -form ω = d x ω = d x omega=dx\omega=d xω=dx. What's the justification for that? Why change a perfectly reasonable (so we were told) infinitesimal change into a more complicated object called a 1-form?
The first thing to point out is that, in terms of doing practical calculus calculations, there's nothing wrong with the traditional treatment of d x i d x i dx^(i)d x^{i}dxi as infinitesimals. It does turn out, though, that mathematicians aren't too happy talking about infinitely small changes; they see the notion as being a bit sloppy. The modern way of interpreting d x i d x i dx^(i)d x^{i}dxi is as functions of tangent vectors, ie as 1-forms. Spivak [21] explains:
Classical differential geometers (and classical analysts) did not hesitate to talk about 'infinitely small' changes d x i d x i dx^(i)d x^{i}dxi of the coordinates x i x i x^(i)x^{i}xi, just as Leibniz had. No one wanted to admit that this was nonsense, because true results were obtained when these infinitely small quantities were divided into each other (provided one did it in the right way).
Eventually it was realized that the closest one can come to describing an infinitely small change is to describe a direction in which this change is supposed to occur, ie a tangent vector. Since d f d f dfd fdf is supposed to be the infinitesimal change of f f fff under an infinitesimal change of the point, d f d f dfd fdf must be a function of this change, which means that d f d f dfd fdf should be a function on tangent vectors. The d x i d x i dx^(i)d x^{i}dxi themselves then metamorphosed into functions ...
The old notation still works, we can still use it, but the underlying machinery has been changed and made more mathematically rigorous. We'll be taking a closer look at the infinitesimal vs 1 -form view of differentials later in this chapter. Our essential takeaway for now is that, depending on context, we can regard the differentials d x i d x i dx^(i)d x^{i}dxi either as infinitesimals or 1-forms.

2.2.2 How this works - basis 1 -forms and basis vectors

So how do differential forms eat tangent vectors and spit out numbers? The answer lies in the relationship between basis vectors and basis 1-forms. Here's a vector:
(2.2.2) v = 3 e ^ x − e ^ y + 6 e ^ z (2.2.2) v = 3 e ^ x − e ^ y + 6 e ^ z {:(2.2.2)v=3 hat(e)_(x)- hat(e)_(y)+6 hat(e)_(z):}\begin{equation*} \mathbf{v}=3 \hat{\mathbf{e}}_{x}-\hat{\mathbf{e}}_{y}+6 \hat{\mathbf{e}}_{z} \tag{2.2.2} \end{equation*}(2.2.2)v=3e^x−e^y+6e^z
with the standard unit basis vectors e ^ x , e ^ y , e ^ z e ^ x , e ^ y , e ^ z hat(e)_(x), hat(e)_(y), hat(e)_(z)\hat{\mathbf{e}}_{x}, \hat{\mathbf{e}}_{y}, \hat{\mathbf{e}}_{z}e^x,e^y,e^z.
And here's a 1-form:
(2.2.3) ω = 5 d x + 2 d y + 4 d z (2.2.3) ω = 5 d x + 2 d y + 4 d z {:(2.2.3)omega=5dx+2dy+4dz:}\begin{equation*} \omega=5 d x+2 d y+4 d z \tag{2.2.3} \end{equation*}(2.2.3)ω=5dx+2dy+4dz
where d x , d y d x , d y dx,dyd x, d ydx,dy and d z d z dzd zdz are the corresponding set of basis 1 -forms. We say 'corresponding' because the basis vectors we choose induce a unique 1 -form basis. (Or vice versa: a set of basis 1-forms will induce a unique set of basis vectors.) Mathematicians are able to define these bases so they act on each other as follows (we'll explore the reasons behind this definition later in this chapter):
(2.2.4) d x ( e ^ x ) = 1 , d x ( e ^ y ) = 0 , d x ( e ^ z ) = 0 d y ( e ^ x ) = 0 , d y ( e ^ y ) = 1 , d y ( e ^ z ) = 0 d z ( e ^ x ) = 0 , d z ( e ^ y ) = 0 , d z ( e ^ z ) = 1 (2.2.4) d x e ^ x = 1 , d x e ^ y = 0 , d x e ^ z = 0 d y e ^ x = 0 , d y e ^ y = 1 , d y e ^ z = 0 d z e ^ x = 0 , d z e ^ y = 0 , d z e ^ z = 1 {:[(2.2.4)dx( hat(e)_(x))=1","dx( hat(e)_(y))=0","dx( hat(e)_(z))=0],[dy( hat(e)_(x))=0","dy( hat(e)_(y))=1","dy( hat(e)_(z))=0],[dz( hat(e)_(x))=0","dz( hat(e)_(y))=0","dz( hat(e)_(z))=1]:}\begin{align*} & d x\left(\hat{\mathbf{e}}_{x}\right)=1, d x\left(\hat{\mathbf{e}}_{y}\right)=0, d x\left(\hat{\mathbf{e}}_{z}\right)=0 \tag{2.2.4}\\ & d y\left(\hat{\mathbf{e}}_{x}\right)=0, d y\left(\hat{\mathbf{e}}_{y}\right)=1, d y\left(\hat{\mathbf{e}}_{z}\right)=0 \\ & d z\left(\hat{\mathbf{e}}_{x}\right)=0, d z\left(\hat{\mathbf{e}}_{y}\right)=0, d z\left(\hat{\mathbf{e}}_{z}\right)=1 \end{align*}(2.2.4)dx(e^x)=1,dx(e^y)=0,dx(e^z)=0dy(e^x)=0,dy(e^y)=1,dy(e^z)=0dz(e^x)=0,dz(e^y)=0,dz(e^z)=1
The basis vectors and basis 1-forms are said to be dual to each other - one acts on the other to produce either 1 or 0 . If you're familiar with the Kronecker delta and index notation, this dual relationship can be succinctly written as
d x i ( e j ) = δ j i = { 1 if i = j 0 if i ≠ j . d x i e j = δ j i = 1       if  i = j 0       if  i ≠ j . dx^(i)(e_(j))=delta_(j)^(i)={[1," if "i=j],[0," if "i!=j].:}d x^{i}\left(\mathbf{e}_{j}\right)=\delta_{j}^{i}=\left\{\begin{array}{ll} 1 & \text { if } i=j \\ 0 & \text { if } i \neq j \end{array} .\right.dxi(ej)=δji={1 if i=j0 if i≠j.
So, at a point, when the 1 -form (2.2.3) acts on the vector (2.2.2) we can write
ω ( v ) = ( 5 d x + 2 d y + 4 d z ) ( 3 e ^ x − e ^ y + 6 e ^ z ) = ( 5 × 3 ) + ( 2 × − 1 ) + ( 4 × 6 ) = 37 ω ( v ) = ( 5 d x + 2 d y + 4 d z ) 3 e ^ x − e ^ y + 6 e ^ z = ( 5 × 3 ) + ( 2 × − 1 ) + ( 4 × 6 ) = 37 {:[omega(v)=(5dx+2dy+4dz)(3 hat(e)_(x)- hat(e)_(y)+6 hat(e)_(z))],[=(5xx3)+(2xx-1)+(4xx6)],[=37]:}\begin{gathered} \omega(\mathbf{v})=(5 d x+2 d y+4 d z)\left(3 \hat{\mathbf{e}}_{x}-\hat{\mathbf{e}}_{y}+6 \hat{\mathbf{e}}_{z}\right) \\ =(5 \times 3)+(2 \times-1)+(4 \times 6) \\ =37 \end{gathered}ω(v)=(5dx+2dy+4dz)(3e^x−e^y+6e^z)=(5×3)+(2×−1)+(4×6)=37
Effectively, we're multiplying the 1 -form components by the respective vector components, ie the 1 -form ω = a d x + b d y + c d z ω = a d x + b d y + c d z omega=adx+bdy+cdz\omega=a d x+b d y+c d zω=adx+bdy+cdz acts on a vector v = d e ^ x + e e ^ y + f e ^ z v = d e ^ x + e e ^ y + f e ^ z v=d hat(e)_(x)+e hat(e)_(y)+f hat(e)_(z)\mathbf{v}=d \hat{\mathbf{e}}_{x}+e \hat{\mathbf{e}}_{y}+f \hat{\mathbf{e}}_{z}v=de^x+ee^y+fe^z as:
ω ( v ) = a d + b e + c f ω ( v ) = a d + b e + c f omega(v)=ad+be+cf\omega(\mathbf{v})=a d+b e+c fω(v)=ad+be+cf
which, of course, is a number. Using index notation, we write
ω ( v ) = ω i v i ω ( v ) = ω i v i omega(v)=omega_(i)v^(i)\omega(\mathbf{v})=\omega_{i} v^{i}ω(v)=ωivi
We won't go there, but using the coordinate transformation rules for 1-forms and tangent vectors (also known as the covariant and contravariant transformation rules) it is straightforward to show that:
  • ω ( v ) ω ( v ) omega(v)\omega(\mathbf{v})ω(v) is an invariant quantity.
If we used those rules to change to a different coordinate system, spherical coordinates ( r , θ , ϕ ) ( r , θ , ϕ ) (r,theta,phi)(r, \theta, \phi)(r,θ,ϕ), for example, ω = 5 d x + 2 d y + 4 d z ω = 5 d x + 2 d y + 4 d z omega=5dx+2dy+4dz\omega=5 d x+2 d y+4 d zω=5dx+2dy+4dz and v = 3 e ^ x − e ^ y + 6 e ^ z v = 3 e ^ x − e ^ y + 6 e ^ z v=3 hat(e)_(x)- hat(e)_(y)+6 hat(e)_(z)\mathbf{v}=3 \hat{\mathbf{e}}_{x}-\hat{\mathbf{e}}_{y}+6 \hat{\mathbf{e}}_{z}v=3e^x−e^y+6e^z would have different components and different bases but ω ( v ) ω ( v ) omega(v)\omega(\mathbf{v})ω(v) would still be 37 .
We mentioned in section 1.2 that the set of tangent vectors at a point P P PPP is a vector space called the tangent space T P M T P M T_(P)MT_{P} MTPM of the manifold M M MMM at P P PPP. The set of 1-forms at P P PPP
is also a vector space (of the same dimension as T P M T P M T_(P)MT_{P} MTPM and M M MMM ) that goes by the name of the cotangent space T P ∗ M T P ∗ M T_(P)^(**)MT_{P}^{*} MTP∗M at P P PPP. In maths-speak, T P ∗ M T P ∗ M T_(P)^(**)MT_{P}^{*} MTP∗M is defined to be the dual space of T P M T P M T_(P)MT_{P} MTPM, meaning the basis vectors from T P M T P M T_(P)MT_{P} MTPM and basis 1-forms from T P ∗ M T P ∗ M T_(P)^(**)MT_{P}^{*} MTP∗M act on each other as described above to give either 1 or 0 .
  • In other words, 1-forms are functions on the tangent space, ie they have a domain (the set of their possible input values) of T P M T P M T_(P)MT_{P} MTPM and a range (the set of their possible outputs) of the real numbers. Using slightly more advanced mathematical language, this dual relationship can be written as
ω P : T P M → R ω P : T P M → R omega_(P):T_(P)M rarrR\omega_{P}: T_{P} M \rightarrow \mathbb{R}ωP:TPM→R
meaning ω P ω P omega_(P)\omega_{P}ωP (an element of the cotangent space T P ∗ M T P ∗ M T_(P)^(**)MT_{P}^{*} MTP∗M, ie a 1-form) acts on an element of the tangent space T P M T P M T_(P)MT_{P} MTPM (ie a tangent vector) to give a real number. Similarly, k k kkk-forms (where k > 1 k > 1 k > 1k>1k>1 ) have a domain of multiple copies of T P M T P M T_(P)MT_{P} MTPM and, again, a range of the real numbers.

2.2.3 d x i d x i dx^(i)d x^{i}dxi picks out a vector's i i iii th component

If the basis 1-form d x d x dxd xdx acts, at a point, on the vector v = 3 e ^ x − e ^ y + 6 e ^ z v = 3 e ^ x − e ^ y + 6 e ^ z v=3 hat(e)_(x)- hat(e)_(y)+6 hat(e)_(z)\mathbf{v}=3 \hat{\mathbf{e}}_{x}-\hat{\mathbf{e}}_{y}+6 \hat{\mathbf{e}}_{z}v=3e^x−e^y+6e^z it picks out the vector's e ^ x e ^ x hat(e)_(x)\hat{\mathbf{e}}_{x}e^x component, which in this case is 3 :
ω ( v ) = d x ( 3 e ^ x − e ^ y + 6 e ^ z ) = ( 1 × 3 ) − 0 + 0 = 3 ω ( v ) = d x 3 e ^ x − e ^ y + 6 e ^ z = ( 1 × 3 ) − 0 + 0 = 3 {:[omega(v)=dx(3 hat(e)_(x)- hat(e)_(y)+6 hat(e)_(z))],[=(1xx3)-0+0],[=3]:}\begin{gathered} \omega(\mathbf{v})=d x\left(3 \hat{\mathbf{e}}_{x}-\hat{\mathbf{e}}_{y}+6 \hat{\mathbf{e}}_{z}\right) \\ =(1 \times 3)-0+0 \\ =3 \end{gathered}ω(v)=dx(3e^x−e^y+6e^z)=(1×3)−0+0=3
because d x ( e ^ x ) = 1 , d x ( e ^ y ) = 0 d x e ^ x = 1 , d x e ^ y = 0 dx( hat(e)_(x))=1,dx( hat(e)_(y))=0d x\left(\hat{\mathbf{e}}_{x}\right)=1, d x\left(\hat{\mathbf{e}}_{y}\right)=0dx(e^x)=1,dx(e^y)=0 and d x ( e ^ z ) = 0 d x e ^ z = 0 dx( hat(e)_(z))=0d x\left(\hat{\mathbf{e}}_{z}\right)=0dx(e^z)=0.
Using index notation we can say
(2.2.5) d x i ( v ) = v i (2.2.5) d x i ( v ) = v i {:(2.2.5)dx^(i)(v)=v^(i):}\begin{equation*} d x^{i}(\mathbf{v})=v^{i} \tag{2.2.5} \end{equation*}(2.2.5)dxi(v)=vi
In other words, d x i d x i dx^(i)d x^{i}dxi picks out the i i iii th component of a vector v v v\mathbf{v}v.
Geometrically, we can interpret (2.2.5) as giving the projection of v v v\mathbf{v}v onto the x i x i x^(i)x^{i}xi th coordinate axis. So d x ( v ) = 3 d x ( v ) = 3 dx(v)=3d x(\mathbf{v})=3dx(v)=3 gives the projection of v v v\mathbf{v}v onto the x x xxx axis; d y ( v ) = − 1 d y ( v ) = − 1 dy(v)=-1d y(\mathbf{v})=-1dy(v)=−1 gives the projection of v v v\mathbf{v}v onto the y y yyy axis; and d z ( v ) = 6 d z ( v ) = 6 dz(v)=6d z(\mathbf{v})=6dz(v)=6 gives the projection of v v v\mathbf{v}v onto the z z zzz axis. If we let a d x a d x adxa d xadx (where a a aaa is a constant) act on v , a d x ( v ) = 3 a v , a d x ( v ) = 3 a v,adx(v)=3a\mathbf{v}, a d x(\mathbf{v})=3 av,adx(v)=3a gives the projection of v v v\mathbf{v}v onto the x x xxx axis multiplied by the factor a a aaa.
In a small abuse of notation, we may omit the basis vectors and write
d x ( 3 , − 1 , 6 ) = 3 d x ( 3 , − 1 , 6 ) = 3 dx(3,-1,6)=3d x(3,-1,6)=3dx(3,−1,6)=3
or, for d y d y dyd ydy acting on vector w = 12 e ^ x + 7 e ^ y + 1 e ^ z w = 12 e ^ x + 7 e ^ y + 1 e ^ z w=12 hat(e)_(x)+7 hat(e)_(y)+1 hat(e)_(z)\mathbf{w}=12 \hat{\mathbf{e}}_{x}+7 \hat{\mathbf{e}}_{y}+1 \hat{\mathbf{e}}_{z}w=12e^x+7e^y+1e^z,
d y ( 12 , 7 , 1 ) = 7 d y ( 12 , 7 , 1 ) = 7 dy(12,7,1)=7d y(12,7,1)=7dy(12,7,1)=7

2.2.4 An example using polar coordinates

2.2.4.1 Calculating the basis vectors and 1-forms

Let's see how this works in polar coordinates. We'll start by calculating the polar basis vectors e r , e θ e r , e θ e_(r),e_(theta)\mathbf{e}_{r}, \mathbf{e}_{\theta}er,eθ and the polar basis 1-forms d r , d θ d r , d θ dr,d thetad r, d \thetadr,dθ, and then show that when they act on each other the result is either 1 or 0 . In other words, we want to show:
d r ( e r ) = 1 d θ ( e θ ) = 1 d r ( e θ ) = 0 d θ ( e r ) = 0 d r e r = 1 d θ e θ = 1 d r e θ = 0 d θ e r = 0 {:[dr(e_(r))=1],[d theta(e_(theta))=1],[dr(e_(theta))=0],[d theta(e_(r))=0]:}\begin{aligned} & d r\left(\mathbf{e}_{r}\right)=1 \\ & d \theta\left(\mathbf{e}_{\theta}\right)=1 \\ & d r\left(\mathbf{e}_{\theta}\right)=0 \\ & d \theta\left(\mathbf{e}_{r}\right)=0 \end{aligned}dr(er)=1dθ(eθ)=1dr(eθ)=0dθ(er)=0
First, the basis vectors. In order to proceed, we need the formula for how Cartesian basis vectors transform to polar basis vectors (which we'll derive in section 2.2.6.2). For now, here it is:
(2.2.6) ( e r , e θ ) = ( e ^ x , e ^ y ) [ ∂ x ∂ r ∂ x ∂ θ ∂ y ∂ r ∂ y ∂ θ ] (2.2.6) e r , e θ = e ^ x , e ^ y ∂ x ∂ r ∂ x ∂ θ ∂ y ∂ r ∂ y ∂ θ {:(2.2.6)(e_(r),e_(theta))=( hat(e)_(x), hat(e)_(y))[[(del x)/(del r),(del x)/(del theta)],[(del y)/(del r),(del y)/(del theta)]]:}\left(\mathbf{e}_{r}, \mathbf{e}_{\theta}\right)=\left(\hat{\mathbf{e}}_{x}, \hat{\mathbf{e}}_{y}\right)\left[\begin{array}{cc} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \tag{2.2.6}\\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{array}\right](2.2.6)(er,eθ)=(e^x,e^y)[∂x∂r∂x∂θ∂y∂r∂y∂θ]
Multiplying this out we get
( e r , e θ ) = ( ∂ x ∂ r e ^ x + ∂ y ∂ r e ^ y , ∂ x ∂ θ e ^ x + ∂ y ∂ θ e ^ y ) e r , e θ = ∂ x ∂ r e ^ x + ∂ y ∂ r e ^ y , ∂ x ∂ θ e ^ x + ∂ y ∂ θ e ^ y (e_(r),e_(theta))=((del x)/(del r) hat(e)_(x)+(del y)/(del r) hat(e)_(y),(del x)/(del theta) hat(e)_(x)+(del y)/(del theta) hat(e)_(y))\left(\mathbf{e}_{r}, \mathbf{e}_{\theta}\right)=\left(\frac{\partial x}{\partial r} \hat{\mathbf{e}}_{x}+\frac{\partial y}{\partial r} \hat{\mathbf{e}}_{y}, \frac{\partial x}{\partial \theta} \hat{\mathbf{e}}_{x}+\frac{\partial y}{\partial \theta} \hat{\mathbf{e}}_{y}\right)(er,eθ)=(∂x∂re^x+∂y∂re^y,∂x∂θe^x+∂y∂θe^y)
In polar coordinates, x = r cos ⁡ θ x = r cos ⁡ θ x=r cos thetax=r \cos \thetax=rcos⁡θ and y = r sin ⁡ θ y = r sin ⁡ θ y=r sin thetay=r \sin \thetay=rsin⁡θ. And, going the other way, r = ( x 2 + y 2 ) 1 / 2 r = x 2 + y 2 1 / 2 r=(x^(2)+y^(2))^(1//2)r=\left(x^{2}+y^{2}\right)^{1 / 2}r=(x2+y2)1/2 and θ = arctan ⁡ ( y / x ) θ = arctan ⁡ ( y / x ) theta=arctan(y//x)\theta=\arctan (y / x)θ=arctan⁡(y/x).
The basis vectors are, starting with e r e r e_(r)\mathbf{e}_{r}er,
e r = ∂ x ∂ r e ^ x + ∂ y ∂ r e ^ y e r = cos ⁡ θ e ^ x + sin ⁡ θ e ^ y e r = ∂ x ∂ r e ^ x + ∂ y ∂ r e ^ y e r = cos ⁡ θ e ^ x + sin ⁡ θ e ^ y {:[e_(r)=(del x)/(del r) hat(e)_(x)+(del y)/(del r) hat(e)_(y)],[e_(r)=cos theta hat(e)_(x)+sin theta hat(e)_(y)]:}\begin{gathered} \mathbf{e}_{r}=\frac{\partial x}{\partial r} \hat{\mathbf{e}}_{x}+\frac{\partial y}{\partial r} \hat{\mathbf{e}}_{y} \\ \mathbf{e}_{r}=\cos \theta \hat{\mathbf{e}}_{x}+\sin \theta \hat{\mathbf{e}}_{y} \end{gathered}er=∂x∂re^x+∂y∂re^yer=cos⁡θe^x+sin⁡θe^y
And for e θ e θ e_(theta)\mathbf{e}_{\theta}eθ,
e θ = ∂ x ∂ θ e ^ x + ∂ y ∂ θ e ^ y e θ = − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y e θ = ∂ x ∂ θ e ^ x + ∂ y ∂ θ e ^ y e θ = − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y {:[e_(theta)=(del x)/(del theta) hat(e)_(x)+(del y)/(del theta) hat(e)_(y)],[e_(theta)=-r sin theta hat(e)_(x)+r cos theta hat(e)_(y)]:}\begin{gathered} \mathbf{e}_{\theta}=\frac{\partial x}{\partial \theta} \hat{\mathbf{e}}_{x}+\frac{\partial y}{\partial \theta} \hat{\mathbf{e}}_{y} \\ \mathbf{e}_{\theta}=-r \sin \theta \hat{\mathbf{e}}_{x}+r \cos \theta \hat{\mathbf{e}}_{y} \end{gathered}eθ=∂x∂θe^x+∂y∂θe^yeθ=−rsin⁡θe^x+rcos⁡θe^y
Next, the basis 1-forms. As θ θ theta\thetaθ is a function of x x xxx and y y yyy, we can find the differential d θ d θ d thetad \thetadθ using ( 2.2 .1 ) ( 2.2 .1 ) (2.2.1)(2.2 .1)(2.2.1)
d f = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z d f = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z df=(del f)/(del x)dx+(del f)/(del y)dy+(del f)/(del z)dzd f=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y+\frac{\partial f}{\partial z} d zdf=∂f∂xdx+∂f∂ydy+∂f∂zdz
And we can say
d θ = ∂ θ ∂ x d x + ∂ θ ∂ y d y d θ = ∂ θ ∂ x d x + ∂ θ ∂ y d y d theta=(del theta)/(del x)dx+(del theta)/(del y)dyd \theta=\frac{\partial \theta}{\partial x} d x+\frac{\partial \theta}{\partial y} d ydθ=∂θ∂xdx+∂θ∂ydy
d θ = − 1 r sin ⁡ θ d x + 1 r cos ⁡ θ d y d θ = − 1 r sin ⁡ θ d x + 1 r cos ⁡ θ d y d theta=-(1)/(r)sin theta dx+(1)/(r)cos theta dyd \theta=-\frac{1}{r} \sin \theta d x+\frac{1}{r} \cos \theta d ydθ=−1rsin⁡θdx+1rcos⁡θdy
Similarly, for d r d r drd rdr
d r = cos ⁡ θ d x + sin ⁡ θ d y d r = cos ⁡ θ d x + sin ⁡ θ d y dr=cos theta dx+sin theta dyd r=\cos \theta d x+\sin \theta d ydr=cos⁡θdx+sin⁡θdy
We saw above in (2.2.4) how the Cartesian basis vectors and basis 1-forms are dual to each other. The same is true with our newly minted polar basis vectors e r , e θ e r , e θ e_(r),e_(theta)\mathbf{e}_{r}, \mathbf{e}_{\theta}er,eθ and basis 1 -forms d r , d θ d r , d θ dr,d thetad r, d \thetadr,dθ :
d r ( e r ) = ( cos ⁡ θ d x + sin ⁡ θ d y ) ( cos ⁡ θ e ^ x + sin ⁡ θ e ^ y ) = 1 d r e r = ( cos ⁡ θ d x + sin ⁡ θ d y ) cos ⁡ θ e ^ x + sin ⁡ θ e ^ y = 1 dr(e_(r))=(cos theta dx+sin theta dy)(cos theta hat(e)_(x)+sin theta hat(e)_(y))=1d r\left(\mathbf{e}_{r}\right)=(\cos \theta d x+\sin \theta d y)\left(\cos \theta \hat{\mathbf{e}}_{x}+\sin \theta \hat{\mathbf{e}}_{y}\right)=1dr(er)=(cos⁡θdx+sin⁡θdy)(cos⁡θe^x+sin⁡θe^y)=1
because cos 2 ⁡ θ + sin 2 ⁡ θ = 1 cos 2 ⁡ θ + sin 2 ⁡ θ = 1 cos^(2)theta+sin^(2)theta=1\cos ^{2} \theta+\sin ^{2} \theta=1cos2⁡θ+sin2⁡θ=1. Similarly
d θ ( e θ ) = ( − 1 r sin ⁡ θ d x + 1 r cos ⁡ θ d y ) ( − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y ) = 1 d r ( e θ ) = ( cos ⁡ θ d x + sin ⁡ θ d y ) ( − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y ) = 0 d θ ( e r ) = ( − 1 r sin ⁡ θ d x + 1 r cos ⁡ θ d y ) ( cos ⁡ θ e ^ x + sin ⁡ θ e ^ y ) = 0 d θ e θ = − 1 r sin ⁡ θ d x + 1 r cos ⁡ θ d y − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y = 1 d r e θ = ( cos ⁡ θ d x + sin ⁡ θ d y ) − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y = 0 d θ e r = − 1 r sin ⁡ θ d x + 1 r cos ⁡ θ d y cos ⁡ θ e ^ x + sin ⁡ θ e ^ y = 0 {:[d theta(e_(theta))=(-(1)/(r)sin theta dx+(1)/(r)cos theta dy)(-r sin theta hat(e)_(x)+r cos theta hat(e)_(y))=1],[dr(e_(theta))=(cos theta dx+sin theta dy)(-r sin theta hat(e)_(x)+r cos theta hat(e)_(y))=0],[d theta(e_(r))=(-(1)/(r)sin theta dx+(1)/(r)cos theta dy)(cos theta hat(e)_(x)+sin theta hat(e)_(y))=0]:}\begin{gathered} d \theta\left(\mathbf{e}_{\theta}\right)=\left(-\frac{1}{r} \sin \theta d x+\frac{1}{r} \cos \theta d y\right)\left(-r \sin \theta \hat{\mathbf{e}}_{x}+r \cos \theta \hat{\mathbf{e}}_{y}\right)=1 \\ d r\left(\mathbf{e}_{\theta}\right)=(\cos \theta d x+\sin \theta d y)\left(-r \sin \theta \hat{\mathbf{e}}_{x}+r \cos \theta \hat{\mathbf{e}}_{y}\right)=0 \\ d \theta\left(\mathbf{e}_{r}\right)=\left(-\frac{1}{r} \sin \theta d x+\frac{1}{r} \cos \theta d y\right)\left(\cos \theta \hat{\mathbf{e}}_{x}+\sin \theta \hat{\mathbf{e}}_{y}\right)=0 \end{gathered}dθ(eθ)=(−1rsin⁡θdx+1rcos⁡θdy)(−rsin⁡θe^x+rcos⁡θe^y)=1dr(eθ)=(cos⁡θdx+sin⁡θdy)(−rsin⁡θe^x+rcos⁡θe^y)=0dθ(er)=(−1rsin⁡θdx+1rcos⁡θdy)(cos⁡θe^x+sin⁡θe^y)=0
Note that unlike their Cartesian counterparts, the polar basis vectors e r , e θ e r , e θ e_(r),e_(theta)\mathbf{e}_{r}, \mathbf{e}_{\theta}er,eθ and basis 1 -forms d r , d θ d r , d θ dr,d thetad r, d \thetadr,dθ change from one point to another.
We can also easily verify that although e r e r e_(r)\mathbf{e}_{r}er is a unit vector, e θ e θ e_(theta)\mathbf{e}_{\theta}eθ is not. To show this, we change e r , e θ e r , e θ e_(r),e_(theta)\mathbf{e}_{r}, \mathbf{e}_{\theta}er,eθ to polar unit vectors e ^ r , e ^ θ e ^ r , e ^ θ hat(e)_(r), hat(e)_(theta)\hat{\mathbf{e}}_{r}, \hat{\mathbf{e}}_{\theta}e^r,e^θ by dividing by their magnitudes:
e ^ r = e r ‖ e r ‖ = cos ⁡ θ e ^ x + sin ⁡ θ e ^ y ‖ cos ⁡ θ e ^ x + sin ⁡ θ e ^ y ‖ = cos ⁡ θ e ^ x + sin ⁡ θ e ^ y cos 2 ⁡ θ + sin 2 ⁡ θ = cos ⁡ θ e ^ x + sin ⁡ θ e ^ y = e r e ^ θ = e θ ‖ e θ ‖ = − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y ‖ − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y ‖ = − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y r ( sin 2 ⁡ θ + cos 2 ⁡ θ ) = − sin ⁡ θ e ^ x + cos ⁡ θ e ^ y = e θ r ≠ e θ e ^ r = e r e r = cos ⁡ θ e ^ x + sin ⁡ θ e ^ y cos ⁡ θ e ^ x + sin ⁡ θ e ^ y = cos ⁡ θ e ^ x + sin ⁡ θ e ^ y cos 2 ⁡ θ + sin 2 ⁡ θ = cos ⁡ θ e ^ x + sin ⁡ θ e ^ y = e r e ^ θ = e θ e θ = − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y = − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y r sin 2 ⁡ θ + cos 2 ⁡ θ = − sin ⁡ θ e ^ x + cos ⁡ θ e ^ y = e θ r ≠ e θ {:[ hat(e)_(r)=(e_(r))/(||e_(r)||)=(cos theta hat(e)_(x)+sin theta hat(e)_(y))/(||cos theta hat(e)_(x)+sin theta hat(e)_(y)||)=(cos theta hat(e)_(x)+sin theta hat(e)_(y))/(sqrt(cos^(2)theta+sin^(2)theta))],[=cos theta hat(e)_(x)+sin theta hat(e)_(y)=e_(r)],[ hat(e)_(theta)=(e_(theta))/(||e_(theta)||)=(-r sin theta hat(e)_(x)+r cos theta hat(e)_(y))/(||-r sin theta hat(e)_(x)+r cos theta hat(e)_(y)||)=(-r sin theta hat(e)_(x)+r cos theta hat(e)_(y))/(r(sqrt(sin^(2)theta+cos^(2)theta)))],[=-sin theta hat(e)_(x)+cos theta hat(e)_(y)=(e_(theta))/(r)!=e_(theta)]:}\begin{gathered} \hat{\mathbf{e}}_{r}=\frac{\mathbf{e}_{r}}{\left\|\mathbf{e}_{r}\right\|}=\frac{\cos \theta \hat{\mathbf{e}}_{x}+\sin \theta \hat{\mathbf{e}}_{y}}{\left\|\cos \theta \hat{\mathbf{e}}_{x}+\sin \theta \hat{\mathbf{e}}_{y}\right\|}=\frac{\cos \theta \hat{\mathbf{e}}_{x}+\sin \theta \hat{\mathbf{e}}_{y}}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}} \\ =\cos \theta \hat{\mathbf{e}}_{x}+\sin \theta \hat{\mathbf{e}}_{y}=\mathbf{e}_{r} \\ \hat{\mathbf{e}}_{\theta}=\frac{\mathbf{e}_{\theta}}{\left\|\mathbf{e}_{\theta}\right\|}=\frac{-r \sin \theta \hat{\mathbf{e}}_{x}+r \cos \theta \hat{\mathbf{e}}_{y}}{\left\|-r \sin \theta \hat{\mathbf{e}}_{x}+r \cos \theta \hat{\mathbf{e}}_{y}\right\|}=\frac{-r \sin \theta \hat{\mathbf{e}}_{x}+r \cos \theta \hat{\mathbf{e}}_{y}}{r\left(\sqrt{\sin ^{2} \theta+\cos ^{2} \theta}\right)} \\ =-\sin \theta \hat{\mathbf{e}}_{x}+\cos \theta \hat{\mathbf{e}}_{y}=\frac{\mathbf{e}_{\theta}}{r} \neq \mathbf{e}_{\theta} \end{gathered}e^r=er‖er‖=cos⁡θe^x+sin⁡θe^y‖cos⁡θe^x+sin⁡θe^y‖=cos⁡θe^x+sin⁡θe^ycos2⁡θ+sin2⁡θ=cos⁡θe^x+sin⁡θe^y=ere^θ=eθ‖eθ‖=−rsin⁡θe^x+rcos⁡θe^y‖−rsin⁡θe^x+rcos⁡θe^y‖=−rsin⁡θe^x+rcos⁡θe^yr(sin2⁡θ+cos2⁡θ)=−sin⁡θe^x+cos⁡θe^y=eθr≠eθ

2.2.4.2 Tangent vectors to a circle

Figure 2.1: Graph of r = 2 sin ⁡ θ r = 2 sin ⁡ θ r=2sin thetar=2 \sin \thetar=2sin⁡θ.
Recall ( 2.2 .5 ) ( 2.2 .5 ) (2.2.5)(2.2 .5)(2.2.5)
d x i ( v ) = v i d x i ( v ) = v i dx^(i)(v)=v^(i)d x^{i}(\mathbf{v})=v^{i}dxi(v)=vi
which tells us that d x i d x i dx^(i)d x^{i}dxi picks out the i i iii th component of vector v v v\mathbf{v}v. So, for example, i = 1 → d x 1 i = 1 → d x 1 i=1rarr dx^(1)i=1 \rightarrow d x^{1}i=1→dx1 would pick out a vector's first component, and i = 2 → d x 2 i = 2 → d x 2 i=2rarr dx^(2)i=2 \rightarrow d x^{2}i=2→dx2 would pick out a vector's second component. (Remember, the superscripts 1 and 2 are coordinate indices, not exponents.) Let's now try this in polar coordinates with tangent vectors to a circle. We'll pick out the first and second components of tangent vector v ( t ) v ( t ) v(t)\mathbf{v}(t)v(t) using d x 1 = d r d x 1 = d r dx^(1)=drd x^{1}=d rdx1=dr and d x 2 = d θ d x 2 = d θ dx^(2)=d thetad x^{2}=d \thetadx2=dθ. Again, we'll be making use of the Pythagorean identity cos 2 ⁡ θ + sin 2 ⁡ θ = 1 cos 2 ⁡ θ + sin 2 ⁡ θ = 1 cos^(2)theta+sin^(2)theta=1\cos ^{2} \theta+\sin ^{2} \theta=1cos2⁡θ+sin2⁡θ=1.
Figure 2.1 shows the polar graph of the circle r = 2 sin ⁡ θ r = 2 sin ⁡ θ r=2sin thetar=2 \sin \thetar=2sin⁡θ. We can parameterise this circle by letting θ = t θ = t theta=t\theta=tθ=t and r = 2 sin ⁡ t r = 2 sin ⁡ t r=2sin tr=2 \sin tr=2sin⁡t for 0 ≤ t ≤ π 0 ≤ t ≤ π 0 <= t <= pi0 \leq t \leq \pi0≤t≤π. A tangent vector v ( t ) v ( t ) v(t)\mathbf{v}(t)v(t) to the circle is then given by
v = d r d t e r + d θ d t e θ v = ( 2 cos ⁡ t ) e r + 1 e θ v = d r d t e r + d θ d t e θ v = ( 2 cos ⁡ t ) e r + 1 e θ {:[v=(dr)/(dt)e_(r)+(d theta)/(dt)e_(theta)],[v=(2cos t)e_(r)+1e_(theta)]:}\begin{gathered} \mathbf{v}=\frac{d r}{d t} \mathbf{e}_{r}+\frac{d \theta}{d t} \mathbf{e}_{\theta} \\ \mathbf{v}=(2 \cos t) \mathbf{e}_{r}+1 \mathbf{e}_{\theta} \end{gathered}v=drdter+dθdteθv=(2cos⁡t)er+1eθ
Or, in component form, v = ( 2 cos ⁡ t , 1 ) v = ( 2 cos ⁡ t , 1 ) v=(2cos t,1)\mathbf{v}=(2 \cos t, 1)v=(2cos⁡t,1).
Substituting the previously calculated equations for e r e r e_(r)\mathbf{e}_{r}er and e θ e θ e_(theta)\mathbf{e}_{\theta}eθ gives
v = 2 cos ⁡ t ( cos ⁡ θ e ^ x + sin ⁡ θ e ^ y ) + 1 × ( − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y ) v = 2 cos ⁡ t cos ⁡ θ e ^ x + sin ⁡ θ e ^ y + 1 × − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y v=2cos t(cos theta hat(e)_(x)+sin theta hat(e)_(y))+1xx(-r sin theta hat(e)_(x)+r cos theta hat(e)_(y))\mathbf{v}=2 \cos t\left(\cos \theta \hat{\mathbf{e}}_{x}+\sin \theta \hat{\mathbf{e}}_{y}\right)+1 \times\left(-r \sin \theta \hat{\mathbf{e}}_{x}+r \cos \theta \hat{\mathbf{e}}_{y}\right)v=2cos⁡t(cos⁡θe^x+sin⁡θe^y)+1×(−rsin⁡θe^x+rcos⁡θe^y)
Now apply d x 1 = d r = cos ⁡ θ d x + sin ⁡ θ d y d x 1 = d r = cos ⁡ θ d x + sin ⁡ θ d y dx^(1)=dr=cos theta dx+sin theta dyd x^{1}=d r=\cos \theta d x+\sin \theta d ydx1=dr=cos⁡θdx+sin⁡θdy
d r ( v ) = ( cos ⁡ θ d x + sin ⁡ θ d y ) ( 2 cos ⁡ t ( cos ⁡ θ e ^ x + sin ⁡ θ e ^ y ) + ( − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y ) ) = 2 cos ⁡ t d r ( v ) = ( cos ⁡ θ d x + sin ⁡ θ d y ) 2 cos ⁡ t cos ⁡ θ e ^ x + sin ⁡ θ e ^ y + − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y = 2 cos ⁡ t {:[dr(v)=(cos theta dx+sin theta dy)(2cos t(cos theta hat(e)_(x)+sin theta hat(e)_(y))+(-r sin theta hat(e)_(x)+r cos theta hat(e)_(y)))],[=2cos t]:}\begin{gathered} d r(\mathbf{v})=(\cos \theta d x+\sin \theta d y)\left(2 \cos t\left(\cos \theta \hat{\mathbf{e}}_{x}+\sin \theta \hat{\mathbf{e}}_{y}\right)+\left(-r \sin \theta \hat{\mathbf{e}}_{x}+r \cos \theta \hat{\mathbf{e}}_{y}\right)\right) \\ =2 \cos t \end{gathered}dr(v)=(cos⁡θdx+sin⁡θdy)(2cos⁡t(cos⁡θe^x+sin⁡θe^y)+(−rsin⁡θe^x+rcos⁡θe^y))=2cos⁡t
which is the first component of v ( t ) v ( t ) v(t)\mathbf{v}(t)v(t).
Next, apply d x 2 = d θ = − 1 r sin ⁡ θ d x + 1 r cos ⁡ θ d y d x 2 = d θ = − 1 r sin ⁡ θ d x + 1 r cos ⁡ θ d y dx^(2)=d theta=-(1)/(r)sin theta dx+(1)/(r)cos theta dyd x^{2}=d \theta=-\frac{1}{r} \sin \theta d x+\frac{1}{r} \cos \theta d ydx2=dθ=−1rsin⁡θdx+1rcos⁡θdy
d r ( v ) = ( − 1 r sin ⁡ θ d x + 1 r cos ⁡ θ d y ) ( 2 cos ⁡ t ( cos ⁡ θ e ^ x + sin ⁡ θ e ^ y ) + ( − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y ) ) = 1 , d r ( v ) = − 1 r sin ⁡ θ d x + 1 r cos ⁡ θ d y 2 cos ⁡ t cos ⁡ θ e ^ x + sin ⁡ θ e ^ y + − r sin ⁡ θ e ^ x + r cos ⁡ θ e ^ y = 1 , {:[dr(v)=(-(1)/(r)sin theta dx+(1)/(r)cos theta dy)(2cos t(cos theta hat(e)_(x)+sin theta hat(e)_(y))+(-r sin theta hat(e)_(x)+r cos theta hat(e)_(y)))],[=1","]:}\begin{gathered} d r(\mathbf{v})=\left(-\frac{1}{r} \sin \theta d x+\frac{1}{r} \cos \theta d y\right)\left(2 \cos t\left(\cos \theta \hat{\mathbf{e}}_{x}+\sin \theta \hat{\mathbf{e}}_{y}\right)+\left(-r \sin \theta \hat{\mathbf{e}}_{x}+r \cos \theta \hat{\mathbf{e}}_{y}\right)\right) \\ =1, \end{gathered}dr(v)=(−1rsin⁡θdx+1rcos⁡θdy)(2cos⁡t(cos⁡θe^x+sin⁡θe^y)+(−rsin⁡θe^x+rcos⁡θe^y))=1,
which is the second component of v ( t ) v ( t ) v(t)\mathbf{v}(t)v(t).

2.2.5 Curves

A curve is a one-dimensional manifold and can be parameterised by a single parameter, t t ttt for example. So in R 3 R 3 R^(3)\mathbb{R}^{3}R3 we might have a curve described by the parametric equation
Φ ( t ) = ( t , 3 t 2 , 5 t ) Φ ( t ) = t , 3 t 2 , 5 t Phi(t)=(t,3t^(2),5t)\boldsymbol{\Phi}(t)=\left(t, 3 t^{2}, 5 t\right)Φ(t)=(t,3t2,5t)
which means
x = t , y = 3 t 2 , z = 5 t x = t , y = 3 t 2 , z = 5 t x=t,y=3t^(2),z=5tx=t, y=3 t^{2}, z=5 tx=t,y=3t2,z=5t
The tangent vector along this curve is given by
d Φ d t = ( d x d t , d y d t , d z d t ) = ( 1 , 6 t , 5 ) d Φ d t = d x d t , d y d t , d z d t = ( 1 , 6 t , 5 ) (d Phi)/(dt)=((dx)/(dt),(dy)/(dt),(dz)/(dt))=(1,6t,5)\frac{d \boldsymbol{\Phi}}{d t}=\left(\frac{d x}{d t}, \frac{d y}{d t}, \frac{d z}{d t}\right)=(1,6 t, 5)dΦdt=(dxdt,dydt,dzdt)=(1,6t,5)
We can show a 1-form ω ω omega\omegaω acting on d Φ d t d Φ d t (d Phi)/(dt)\frac{d \Phi}{d t}dΦdt by
ω ( d Φ d t ) ω d Φ d t omega((d Phi)/(dt))\omega\left(\frac{d \Phi}{d t}\right)ω(dΦdt)
Equation ( 2.2 .5 ) ( 2.2 .5 ) (2.2.5)(2.2 .5)(2.2.5)
d x i ( v ) = v i d x i ( v ) = v i dx^(i)(v)=v^(i)d x^{i}(\mathbf{v})=v^{i}dxi(v)=vi
tells us that d x i d x i dx^(i)d x^{i}dxi picks out the i i iii th component of vector v v v\mathbf{v}v. So if ω = d y ω = d y omega=dy\omega=d yω=dy, then
d y ( d Φ d t ) = d y ( d x d t , d y d t , d z d t ) = d y ( 1 , 6 t , 5 ) = 6 t d y d Φ d t = d y d x d t , d y d t , d z d t = d y ( 1 , 6 t , 5 ) = 6 t dy((d Phi)/(dt))=dy((dx)/(dt),(dy)/(dt),(dz)/(dt))=dy(1,6t,5)=6td y\left(\frac{d \Phi}{d t}\right)=d y\left(\frac{d x}{d t}, \frac{d y}{d t}, \frac{d z}{d t}\right)=d y(1,6 t, 5)=6 tdy(dΦdt)=dy(dxdt,dydt,dzdt)=dy(1,6t,5)=6t

2.2.6 The differential d f d f dfd fdf revisited

Figure 2.2: Ant on a hotplate.
Say we have a scalar field f f fff on a manifold, ie a function that assigns a scalar (a number) to every point on the manifold. At a given point, the 1 -form d f d f dfd fdf acts on a tangent vector v = d Φ d t v = d Φ d t v=(d Phi)/(dt)\mathbf{v}=\frac{d \Phi}{d t}v=dΦdt to give a number, the directional derivative d f ( v ) d f ( v ) df(v)d f(\mathbf{v})df(v), which tells us how much f f fff is changing in the direction of v v v\mathbf{v}v with respect to t t ttt.
For example, consider a heat-resistant ant on a hotplate walking along a curve C C CCC, as shown in Figure 2.2. Let's assume the temperature of the hotplate is given by the scalar function f ( x , y ) = 100 − ( x 2 + y 2 ) f ( x , y ) = 100 − x 2 + y 2 f(x,y)=100-(x^(2)+y^(2))f(x, y)=100-\left(x^{2}+y^{2}\right)f(x,y)=100−(x2+y2). In the diagram, contours of constant temperature are shown by broken-line circles. We'll also assume that we can describe the ant's progress along C C CCC using some parametric equation Φ ( t ) Φ ( t ) Phi(t)\Phi(t)Φ(t), and that t t ttt measures time in minutes. The tangent vector along C C CCC is given by
v = d Φ d t = ( d x d t , d y d t ) v = d Φ d t = d x d t , d y d t v=(d Phi)/(dt)=((dx)/(dt),(dy)/(dt))\mathbf{v}=\frac{d \boldsymbol{\Phi}}{d t}=\left(\frac{d x}{d t}, \frac{d y}{d t}\right)v=dΦdt=(dxdt,dydt)
As the ant moves along C C CCC, the directional derivative d f ( v ) d f ( v ) df(v)d f(\mathbf{v})df(v) gives the instantaneous rate of temperature change per minute in her direction of travel. So, for example, if at point P ( 3 , 2 ) P ( 3 , 2 ) P(3,2)P(3,2)P(3,2) the tangent vector v = ( 1 , 2 ) v = ( 1 , 2 ) v=(1,2)\mathbf{v}=(1,2)v=(1,2), then
d f ( v ) = − ( 2 x d x + 2 y d y ) ( d x d t , d y d t ) = − ( 6 d x + 4 d y ) ( 1 , 2 ) = − 14 d f ( v ) = − ( 2 x d x + 2 y d y ) d x d t , d y d t = − ( 6 d x + 4 d y ) ( 1 , 2 ) = − 14 df(v)=-(2xdx+2ydy)((dx)/(dt),(dy)/(dt))=-(6dx+4dy)(1,2)=-14d f(\mathbf{v})=-(2 x d x+2 y d y)\left(\frac{d x}{d t}, \frac{d y}{d t}\right)=-(6 d x+4 d y)(1,2)=-14df(v)=−(2xdx+2ydy)(dxdt,dydt)=−(6dx+4dy)(1,2)=−14
Meaning that at P ( 3 , 2 ) P ( 3 , 2 ) P(3,2)P(3,2)P(3,2) the ant would experience a temperature rate of change with respect to time of -14 temperature units per minute in the direction of v v v\mathbf{v}v.

2.2.6.1 d f ( v ) d f ( v ) df(v)d f(\mathbf{v})df(v) as an approximation to Δ f Δ f Delta f\Delta fΔf

Recall from section 2.2.1 the Spivak [21] quote stating that d f d f dfd fdf should correctly be regarded as a 1 -form rather than an infinitesimal change of f f fff. We can now look at this idea a little more closely.
In elementary calculus the differential of a function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x, y, z)f(x,y,z) is a scalar (or number)
d f = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z d f = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z df=(del f)/(del x)dx+(del f)/(del y)dy+(del f)/(del z)dzd f=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y+\frac{\partial f}{\partial z} d zdf=∂f∂xdx+∂f∂ydy+∂f∂zdz
representing either an infinitesimal change in f f fff associated with infinitesimal changes d x , d y , d z d x , d y , d z dx,dy,dzd x, d y, d zdx,dy,dz, or a small change d f ≈ Δ f d f ≈ Δ f df~~Delta fd f \approx \Delta fdf≈Δf associated with small changes d x = Δ x , d y = d x = Δ x , d y = dx=Delta x,dy=d x=\Delta x, d y=dx=Δx,dy= Δ y , d z = Δ z Δ y , d z = Δ z Delta y,dz=Delta z\Delta y, d z=\Delta zΔy,dz=Δz in the independent variables x , y , z x , y , z x,y,zx, y, zx,y,z. So, for example, the above 'hotplate' function f ( x , y ) = 100 − ( x 2 + y 2 ) f ( x , y ) = 100 − x 2 + y 2 f(x,y)=100-(x^(2)+y^(2))f(x, y)=100-\left(x^{2}+y^{2}\right)f(x,y)=100−(x2+y2) gives d f = − ( 2 x d x + 2 y d y ) d f = − ( 2 x d x + 2 y d y ) df=-(2xdx+2ydy)d f=-(2 x d x+2 y d y)df=−(2xdx+2ydy), which at point P ( 3 , 2 ) P ( 3 , 2 ) P(3,2)P(3,2)P(3,2) gives d f = − ( 6 d x + 4 d y ) d f = − ( 6 d x + 4 d y ) df=-(6dx+4dy)d f=-(6 d x+4 d y)df=−(6dx+4dy). If we change x x xxx from 3 to 3.1 and y y yyy from 2 to 2.1 , we have d x = Δ x = 0.1 d x = Δ x = 0.1 dx=Delta x=0.1d x=\Delta x=0.1dx=Δx=0.1 and d y = Δ y = 0.1 d y = Δ y = 0.1 dy=Delta y=0.1d y=\Delta y=0.1dy=Δy=0.1, giving
Δ f = ( 100 − ( 3.1 ) 2 − ( 2.1 ) 2 ) − ( 100 − 3 2 − 2 2 ) = − 1.02 Δ f = 100 − ( 3.1 ) 2 − ( 2.1 ) 2 − 100 − 3 2 − 2 2 = − 1.02 Delta f=(100-(3.1)^(2)-(2.1)^(2))-(100-3^(2)-2^(2))=-1.02\Delta f=\left(100-(3.1)^{2}-(2.1)^{2}\right)-\left(100-3^{2}-2^{2}\right)=-1.02Δf=(100−(3.1)2−(2.1)2)−(100−32−22)=−1.02
and
d f = − ( 6 × 0.1 + 4 × 0.1 ) = − 1 d f = − ( 6 × 0.1 + 4 × 0.1 ) = − 1 df=-(6xx0.1+4xx0.1)=-1d f=-(6 \times 0.1+4 \times 0.1)=-1df=−(6×0.1+4×0.1)=−1
If we repeat the exercise with d x = Δ x = 0.01 d x = Δ x = 0.01 dx=Delta x=0.01d x=\Delta x=0.01dx=Δx=0.01 and d y = Δ y = 0.01 d y = Δ y = 0.01 dy=Delta y=0.01d y=\Delta y=0.01dy=Δy=0.01, we find Δ f = − 0.1002 Δ f = − 0.1002 Delta f=-0.1002\Delta f=-0.1002Δf=−0.1002 and d f = − 0.1 d f = − 0.1 df=-0.1d f=-0.1df=−0.1. And with d x = Δ x = 0.001 d x = Δ x = 0.001 dx=Delta x=0.001d x=\Delta x=0.001dx=Δx=0.001 and d y = Δ y = 0.001 d y = Δ y = 0.001 dy=Delta y=0.001d y=\Delta y=0.001dy=Δy=0.001, we find Δ f = − 0.010002 Δ f = − 0.010002 Delta f=-0.010002\Delta f=-0.010002Δf=−0.010002 and d f = − 0.01 d f = − 0.01 df=-0.01d f=-0.01df=−0.01. In other words, the smaller we make d x = Δ x d x = Δ x dx=Delta xd x=\Delta xdx=Δx and d y = Δ y d y = Δ y dy=Delta yd y=\Delta ydy=Δy, the better is the approximation Δ f ≈ d f Δ f ≈ d f Delta f~~df\Delta f \approx d fΔf≈df.
The modern interpretation of d f d f dfd fdf is as a 1-form, not a number. In order to change the 1 -form d f d f dfd fdf to a number we need to feed it a direction, ie a tangent vector. We can demonstrate this using the same function f ( x , y ) = 100 − ( x 2 + y 2 ) f ( x , y ) = 100 − x 2 + y 2 f(x,y)=100-(x^(2)+y^(2))f(x, y)=100-\left(x^{2}+y^{2}\right)f(x,y)=100−(x2+y2), again at point P ( 3 , 2 ) P ( 3 , 2 ) P(3,2)P(3,2)P(3,2). In effect, we will now repeat the above calculation but with a different interpretation as to the meaning of d f d f dfd fdf. A displacement Δ f Δ f Delta f\Delta fΔf from P P PPP along an ordinary Euclidean vector v = ( v 1 , v 2 ) v = v 1 , v 2 v=(v^(1),v^(2))\mathbf{v}=\left(v^{1}, v^{2}\right)v=(v1,v2) can be written as
Δ f = f ( P + ( v 1 , v 2 ) ) − f ( P ) Δ f = f P + v 1 , v 2 − f ( P ) Delta f=f(P+(v^(1),v^(2)))-f(P)\Delta f=f\left(P+\left(v^{1}, v^{2}\right)\right)-f(P)Δf=f(P+(v1,v2))−f(P)
Now let's input a small vector, v = ( 0.1 , 0.1 ) v = ( 0.1 , 0.1 ) v=(0.1,0.1)\mathbf{v}=(0.1,0.1)v=(0.1,0.1), to (again) get
Δ f = f ( P ( 3.1 , 2.1 ) ) − f ( 3 , 2 ) = ( 100 − ( 3.1 2 + 2.1 2 ) ) − ( 100 − ( 3 2 + 2 2 ) ) = − 1.02 Δ f = f ( P ( 3.1 , 2.1 ) ) − f ( 3 , 2 ) = 100 − 3.1 2 + 2.1 2 − 100 − 3 2 + 2 2 = − 1.02 {:[Delta f=f(P(3.1","2.1))-f(3","2)],[=(100-(3.1^(2)+2.1^(2)))-(100-(3^(2)+2^(2)))=-1.02]:}\begin{gathered} \Delta f=f(P(3.1,2.1))-f(3,2) \\ =\left(100-\left(3.1^{2}+2.1^{2}\right)\right)-\left(100-\left(3^{2}+2^{2}\right)\right)=-1.02 \end{gathered}Δf=f(P(3.1,2.1))−f(3,2)=(100−(3.12+2.12))−(100−(32+22))=−1.02
Because we are working in R 2 R 2 R^(2)\mathbb{R}^{2}R2 (the x y x y xyx yxy plane), we can also treat v v v\mathbf{v}v as a tangent vector and calculate d f ( v ) d f ( v ) df(v)d f(\mathbf{v})df(v), which equals
d f ( v ) = − ( 6 d x + 4 d y ) ( 0.1 , 0.1 ) = − 1 d f ( v ) = − ( 6 d x + 4 d y ) ( 0.1 , 0.1 ) = − 1 df(v)=-(6dx+4dy)(0.1,0.1)=-1d f(\mathbf{v})=-(6 d x+4 d y)(0.1,0.1)=-1df(v)=−(6dx+4dy)(0.1,0.1)=−1
As with the previous ' d f d f dfd fdf as a number' calculations, the smaller we make v v v\mathbf{v}v, the better is the approximation Δ f ≈ d f ( v ) Δ f ≈ d f ( v ) Delta f~~df(v)\Delta f \approx d f(\mathbf{v})Δf≈df(v). By considering d f d f dfd fdf as a 1-form, a linear function that acts on tangent vector, we are thus able to recover the elementary interpretation of d f d f dfd fdf as the small change in f f fff associated with small changes in the independent variables x , y , z x , y , z x,y,zx, y, zx,y,z.
Figure 2.3: The differential as an approximation to Δ f Δ f Delta f\Delta fΔf.
The general, multidimensional case for a function z = f ( x 1 , … , x n , ) z = f x 1 , … , x n , z=f(x^(1),dots,x^(n),)z=f\left(x^{1}, \ldots, x^{n},\right)z=f(x1,…,xn,) is shown in Figure 2.3 (from Lee [12]). In the diagram, D D DDD is a region of R n R n R^(n)\mathbb{R}^{n}Rn with coordinates x 1 , … , x n , v = ( v 1 , … , v n ) x 1 , … , x n , v = v 1 , … , v n x^(1),dots,x^(n),v=(v^(1),dots,v^(n))x^{1}, \ldots, x^{n}, \mathbf{v}=\left(v^{1}, \ldots, v^{n}\right)x1,…,xn,v=(v1,…,vn) is a vector in D D DDD, and P P PPP is a point in D D DDD.
As noted above, a displacement Δ f Δ f Delta f\Delta fΔf from P P PPP along v v v\mathbf{v}v can be written as
Δ f = f ( P + ( v 1 , … , v n ) ) − f ( P ) Δ f = f P + v 1 , … , v n − f ( P ) Delta f=f(P+(v^(1),dots,v^(n)))-f(P)\Delta f=f\left(P+\left(v^{1}, \ldots, v^{n}\right)\right)-f(P)Δf=f(P+(v1,…,vn))−f(P)
where f ( P ) f ( P ) f(P)f(P)f(P) is the value of f f fff at point Q Q QQQ and f ( P + ( v 1 , … , v n ) ) f P + v 1 , … , v n f(P+(v^(1),dots,v^(n)))f\left(P+\left(v^{1}, \ldots, v^{n}\right)\right)f(P+(v1,…,vn)) is the value of f f fff at point R R RRR.
In two dimensions, for a function z = f ( x ) z = f ( x ) z=f(x)z=f(x)z=f(x), tangent T T TTT would be a straight line. In three dimensions, for a function z = f ( x , y ) , T z = f ( x , y ) , T z=f(x,y),Tz=f(x, y), Tz=f(x,y),T would be a plane. In more than three dimensions, T T TTT would be a higher dimensional space. The slope of T T TTT is given by the ratio of 'rise' to 'run'. So the 'rise' of T T TTT equals slope multiplied by 'run'. The slope of T T TTT along the x 1 x 1 x^(1)x^{1}x1 axis is ∂ f ∂ x 1 ∂ f ∂ x 1 (del f)/(delx^(1))\frac{\partial f}{\partial x^{1}}∂f∂x1, the 'run' is v 1 v 1 v^(1)v^{1}v1, giving a 'rise' of ∂ f ∂ x 1 ( v 1 ) ∂ f ∂ x 1 v 1 (del f)/(delx^(1))(v^(1))\frac{\partial f}{\partial x^{1}}\left(v^{1}\right)∂f∂x1(v1). The total 'rise' of T T TTT is therefore given by
( ∂ f ∂ x 1 + ∂ f ∂ x 2 , … , + ∂ f ∂ x n ) ( v 1 + v 2 , … , + v n ) ∂ f ∂ x 1 + ∂ f ∂ x 2 , … , + ∂ f ∂ x n v 1 + v 2 , … , + v n ((del f)/(delx^(1))+(del f)/(delx^(2)),dots,+(del f)/(delx^(n)))(v^(1)+v^(2),dots,+v^(n))\left(\frac{\partial f}{\partial x^{1}}+\frac{\partial f}{\partial x^{2}}, \ldots,+\frac{\partial f}{\partial x^{n}}\right)\left(v^{1}+v^{2}, \ldots,+v^{n}\right)(∂f∂x1+∂f∂x2,…,+∂f∂xn)(v1+v2,…,+vn)
which we can rewrite, knowing that d x i d x i dx^(i)d x^{i}dxi picks out the i i iii th component of v v v\mathbf{v}v, as
( ∂ f ∂ x 1 d x 1 + ∂ f ∂ x 2 d x 2 , … , + ∂ f ∂ x n d x n ) ( v ) = d f ( v ) ∂ f ∂ x 1 d x 1 + ∂ f ∂ x 2 d x 2 , … , + ∂ f ∂ x n d x n ( v ) = d f ( v ) ((del f)/(delx^(1))dx^(1)+(del f)/(delx^(2))dx^(2),dots,+(del f)/(delx^(n))dx^(n))(v)=df(v)\left(\frac{\partial f}{\partial x^{1}} d x^{1}+\frac{\partial f}{\partial x^{2}} d x^{2}, \ldots,+\frac{\partial f}{\partial x^{n}} d x^{n}\right)(\mathbf{v})=d f(\mathbf{v})(∂f∂x1dx1+∂f∂x2dx2,…,+∂f∂xndxn)(v)=df(v)
As is clear from the diagram, the approximation Δ f ≈ d f ( v ) Δ f ≈ d f ( v ) Delta f~~df(v)\Delta f \approx d f(\mathbf{v})Δf≈df(v) improves as v v v\mathbf{v}v becomes smaller.
Lee [ 12 ] [ 12 ] [12][12][12] states:
In other words, d f p [ d f d f p [ d f df_(p)[dfd f_{p}[d fdfp[df at point p ] p ] p]p]p] is the linear functional that best approximates f f fff near p … p … p dotsp \ldotsp… The great power of the concept of the differential comes from the fact that we can define d f d f dfd fdf invariantly on any manifold, without resorting to vague arguments involving infinitesimals.

2.2.6.2 Vectors as differential operators

At a more advanced level, the partial derivative operators ∂ ∂ x i ∂ ∂ x i (del)/(delx^(i))\frac{\partial}{\partial x^{i}}∂∂xi may be usefully regarded as basis vectors. So, for example, the standard unit basis vectors e ^ x , e ^ y , e ^ z e ^ x , e ^ y , e ^ z hat(e)_(x), hat(e)_(y), hat(e)_(z)\hat{\mathbf{e}}_{x}, \hat{\mathbf{e}}_{y}, \hat{\mathbf{e}}_{z}e^x,e^y,e^z may be identified with the partial derivative operators ∂ ∂ x , ∂ ∂ y , ∂ ∂ z ∂ ∂ x , ∂ ∂ y , ∂ ∂ z (del)/(del x),(del)/(del y),(del)/(del z)\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}∂∂x,∂∂y,∂∂z. A vector at point P P PPP,
v = v 1 e ^ x + v 2 e ^ y + v 3 e ^ z = v 1 ∂ ∂ x + v 2 ∂ ∂ y + v 3 ∂ ∂ z v = v 1 e ^ x + v 2 e ^ y + v 3 e ^ z = v 1 ∂ ∂ x + v 2 ∂ ∂ y + v 3 ∂ ∂ z v=v^(1) hat(e)_(x)+v^(2) hat(e)_(y)+v^(3) hat(e)_(z)=v^(1)(del)/(del x)+v^(2)(del)/(del y)+v^(3)(del)/(del z)\mathbf{v}=v^{1} \hat{\mathbf{e}}_{x}+v^{2} \hat{\mathbf{e}}_{y}+v^{3} \hat{\mathbf{e}}_{z}=v^{1} \frac{\partial}{\partial x}+v^{2} \frac{\partial}{\partial y}+v^{3} \frac{\partial}{\partial z}v=v1e^x+v2e^y+v3e^z=v1∂∂x+v2∂∂y+v3∂∂z
is then an operator waiting to act on a function f f fff to give a number
v f = v 1 ∂ f ∂ x + v 2 ∂ f ∂ y + v 3 ∂ f ∂ z v f = v 1 ∂ f ∂ x + v 2 ∂ f ∂ y + v 3 ∂ f ∂ z vf=v^(1)(del f)/(del x)+v^(2)(del f)/(del y)+v^(3)(del f)/(del z)\mathbf{v} f=v^{1} \frac{\partial f}{\partial x}+v^{2} \frac{\partial f}{\partial y}+v^{3} \frac{\partial f}{\partial z}vf=v1∂f∂x+v2∂f∂y+v3∂f∂z
the directional derivative at P P PPP of f f fff in the direction of v v v\mathbf{v}v. The differential d f d f dfd fdf of a function f f fff is then defined by d f ( v ) = v f d f ( v ) = v f df(v)=vfd f(\mathbf{v})=\mathbf{v} fdf(v)=vf.
To see how this definition makes sense, we introduce a special type of function called a coordinate function. For example, in R 3 R 3 R^(3)\mathbb{R}^{3}R3 we can think of x , y x , y x,yx, yx,y and z z zzz as Cartesian coordinate functions that act on a point P P PPP to pick out, respectively, the x , y x , y x,yx, yx,y and z z zzz coordinates of P P PPP. So for a point P ( 2 , 7 , 1 ) P ( 2 , 7 , 1 ) P(2,7,1)P(2,7,1)P(2,7,1) we have
x ( P ) = x ( 2 , 7 , 1 ) = 2 y ( P ) = y ( 2 , 7 , 1 ) = 7 z ( P ) = z ( 2 , 7 , 1 ) = 1 x ( P ) = x ( 2 , 7 , 1 ) = 2 y ( P ) = y ( 2 , 7 , 1 ) = 7 z ( P ) = z ( 2 , 7 , 1 ) = 1 {:[x(P)=x(2","7","1)=2],[y(P)=y(2","7","1)=7],[z(P)=z(2","7","1)=1]:}\begin{aligned} & x(P)=x(2,7,1)=2 \\ & y(P)=y(2,7,1)=7 \\ & z(P)=z(2,7,1)=1 \end{aligned}x(P)=x(2,7,1)=2y(P)=y(2,7,1)=7z(P)=z(2,7,1)=1
(We've already met an example of a coordinate function in the shape of the polar to Cartesian conversion formulas x = r cos ⁡ θ x = r cos ⁡ θ x=r cos thetax=r \cos \thetax=rcos⁡θ and y = r sin ⁡ θ y = r sin ⁡ θ y=r sin thetay=r \sin \thetay=rsin⁡θ. If we feed a point P ( r , θ ) P ( r , θ ) P(r,theta)P(r, \theta)P(r,θ) into x = r cos ⁡ θ x = r cos ⁡ θ x=r cos thetax=r \cos \thetax=rcos⁡θ, out pops the x x xxx coordinate of P P PPP, and if we feed the same point into y = r sin ⁡ θ y = r sin ⁡ θ y=r sin thetay=r \sin \thetay=rsin⁡θ, out pops the y y yyy coordinate of P P PPP.)
If we feed the coordinate function x x xxx into d f ( v ) = v f d f ( v ) = v f df(v)=vfd f(\mathbf{v})=\mathbf{v} fdf(v)=vf we get
d x ( v ) = v x = v 1 ∂ x ∂ x + v 2 ∂ x ∂ y + v 3 ∂ x ∂ z d x ( v ) = v x = v 1 ∂ x ∂ x + v 2 ∂ x ∂ y + v 3 ∂ x ∂ z dx(v)=vx=v^(1)(del x)/(del x)+v^(2)(del x)/(del y)+v^(3)(del x)/(del z)d x(\mathbf{v})=\mathbf{v} x=v^{1} \frac{\partial x}{\partial x}+v^{2} \frac{\partial x}{\partial y}+v^{3} \frac{\partial x}{\partial z}dx(v)=vx=v1∂x∂x+v2∂x∂y+v3∂x∂z
The rate of change of the Cartesian coordinate function x x xxx in the direction of x x xxx is 1 , ie ∂ x ∂ x = 1 ∂ x ∂ x = 1 (del x)/(del x)=1\frac{\partial x}{\partial x}=1∂x∂x=1. However, coordinates are independent of each other, so ∂ x ∂ y = ∂ x ∂ z = 0 ∂ x ∂ y = ∂ x ∂ z = 0 (del x)/(del y)=(del x)/(del z)=0\frac{\partial x}{\partial y}=\frac{\partial x}{\partial z}=0∂x∂y=∂x∂z=0. Therefore
d x ( v ) = v x = ( v 1 × 1 ) + ( v 2 × 0 ) + ( v 3 × 0 ) = v 1 d x ( v ) = v x = v 1 × 1 + v 2 × 0 + v 3 × 0 = v 1 dx(v)=vx=(v^(1)xx1)+(v^(2)xx0)+(v^(3)xx0)=v^(1)d x(\mathbf{v})=\mathbf{v} x=\left(v^{1} \times 1\right)+\left(v^{2} \times 0\right)+\left(v^{3} \times 0\right)=v^{1}dx(v)=vx=(v1×1)+(v2×0)+(v3×0)=v1
Similarly, for the coordinate functions y y yyy and z z zzz
d y ( v ) = v y = v 2 d y ( v ) = v y = v 2 dy(v)=vy=v^(2)d y(\mathbf{v})=\mathbf{v} y=v^{2}dy(v)=vy=v2
and
d z ( v ) = v z = v 3 d z ( v ) = v z = v 3 dz(v)=vz=v^(3)d z(\mathbf{v})=\mathbf{v} z=v^{3}dz(v)=vz=v3
And we can see that:
  • the differentials d x , d y d x , d y dx,dyd x, d ydx,dy and d z d z dzd zdz of the Cartesian coordinate functions pick out, respectively, the v 1 , v 2 v 1 , v 2 v^(1),v^(2)v^{1}, v^{2}v1,v2 and v 3 v 3 v^(3)v^{3}v3 components of vector v v v\mathbf{v}v. In other words, they behave in exactly the same way as the basis 1-forms d x , d y d x , d y dx,dyd x, d ydx,dy and d z d z dzd zdz. Hence the notation d x , d y d x , d y dx,dyd x, d ydx,dy and d z d z dzd zdz (more generally, d x i d x i dx^(i)d x^{i}dxi ) for basis 1-forms.
Furthermore, if we set v 1 = v 2 = v 3 = 1 v 1 = v 2 = v 3 = 1 v^(1)=v^(2)=v^(3)=1v^{1}=v^{2}=v^{3}=1v1=v2=v3=1 and let v v v\mathbf{v}v equal, in turn, the basis vectors ∂ ∂ x , ∂ ∂ y , ∂ ∂ z ∂ ∂ x , ∂ ∂ y , ∂ ∂ z (del)/(del x),(del)/(del y),(del)/(del z)\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}∂∂x,∂∂y,∂∂z, we get
d x ( ∂ ∂ x ) = 1 , d x ( ∂ ∂ y ) = 0 , d x ( ∂ ∂ z ) = 0 d y ( ∂ ∂ x ) = 0 , d y ( ∂ ∂ y ) = 1 , d y ( ∂ ∂ z ) = 0 d z ( ∂ ∂ x ) = 0 , d z ( ∂ ∂ y ) = 0 , d z ( ∂ ∂ z ) = 1 d x ∂ ∂ x = 1 , d x ∂ ∂ y = 0 , d x ∂ ∂ z = 0 d y ∂ ∂ x = 0 , d y ∂ ∂ y = 1 , d y ∂ ∂ z = 0 d z ∂ ∂ x = 0 , d z ∂ ∂ y = 0 , d z ∂ ∂ z = 1 {:[dx((del)/(del x))=1","dx((del)/(del y))=0","dx((del)/(del z))=0],[dy((del)/(del x))=0","dy((del)/(del y))=1","dy((del)/(del z))=0],[dz((del)/(del x))=0","dz((del)/(del y))=0","dz((del)/(del z))=1]:}\begin{aligned} & d x\left(\frac{\partial}{\partial x}\right)=1, d x\left(\frac{\partial}{\partial y}\right)=0, d x\left(\frac{\partial}{\partial z}\right)=0 \\ & d y\left(\frac{\partial}{\partial x}\right)=0, d y\left(\frac{\partial}{\partial y}\right)=1, d y\left(\frac{\partial}{\partial z}\right)=0 \\ & d z\left(\frac{\partial}{\partial x}\right)=0, d z\left(\frac{\partial}{\partial y}\right)=0, d z\left(\frac{\partial}{\partial z}\right)=1 \end{aligned}dx(∂∂x)=1,dx(∂∂y)=0,dx(∂∂z)=0dy(∂∂x)=0,dy(∂∂y)=1,dy(∂∂z)=0dz(∂∂x)=0,dz(∂∂y)=0,dz(∂∂z)=1
which is equivalent to, and the justification for, (2.2.4):
d x ( e ^ x ) = 1 , d x ( e ^ y ) = 0 , d x ( e ^ z ) = 0 d y ( e ^ x ) = 0 , d y ( e ^ y ) = 1 , d y ( e ^ z ) = 0 d z ( e ^ x ) = 0 , d z ( e ^ y ) = 0 , d z ( e ^ z ) = 1 d x e ^ x = 1 , d x e ^ y = 0 , d x e ^ z = 0 d y e ^ x = 0 , d y e ^ y = 1 , d y e ^ z = 0 d z e ^ x = 0 , d z e ^ y = 0 , d z e ^ z = 1 {:[dx( hat(e)_(x))=1","dx( hat(e)_(y))=0","dx( hat(e)_(z))=0],[dy( hat(e)_(x))=0","dy( hat(e)_(y))=1","dy( hat(e)_(z))=0],[dz( hat(e)_(x))=0","dz( hat(e)_(y))=0","dz( hat(e)_(z))=1]:}\begin{aligned} & d x\left(\hat{\mathbf{e}}_{x}\right)=1, d x\left(\hat{\mathbf{e}}_{y}\right)=0, d x\left(\hat{\mathbf{e}}_{z}\right)=0 \\ & d y\left(\hat{\mathbf{e}}_{x}\right)=0, d y\left(\hat{\mathbf{e}}_{y}\right)=1, d y\left(\hat{\mathbf{e}}_{z}\right)=0 \\ & d z\left(\hat{\mathbf{e}}_{x}\right)=0, d z\left(\hat{\mathbf{e}}_{y}\right)=0, d z\left(\hat{\mathbf{e}}_{z}\right)=1 \end{aligned}dx(e^x)=1,dx(e^y)=0,dx(e^z)=0dy(e^x)=0,dy(e^y)=1,dy(e^z)=0dz(e^x)=0,dz(e^y)=0,dz(e^z)=1
The basis of partial derivative operators is known as the coordinate basis. Say we have general coordinates x 1 , x 2 , x 3 , … , x n x 1 , x 2 , x 3 , … , x n x^(1),x^(2),x^(3),dots,x^(n)x^{1}, x^{2}, x^{3}, \ldots, x^{n}x1,x2,x3,…,xn, which are not necessarily rectangular or even orthogonal. The coordinate basis vectors for x 1 , x 2 , x 3 , … , x n x 1 , x 2 , x 3 , … , x n x^(1),x^(2),x^(3),dots,x^(n)x^{1}, x^{2}, x^{3}, \ldots, x^{n}x1,x2,x3,…,xn can be understood as follows.
There is a tangent vector that takes x 1 x 1 x^(1)x^{1}x1 to 1 and all the other coordinates to 0 . This tangent vector is the coordinate basis vector e 1 = ∂ ∂ x 1 e 1 = ∂ ∂ x 1 e_(1)=(del)/(delx^(1))\mathbf{e}_{1}=\frac{\partial}{\partial x^{1}}e1=∂∂x1 and has components
∂ ( x 1 , … , x n ) ∂ x 1 = ( 1 , 0 , … , 0 , 0 ) ∂ x 1 , … , x n ∂ x 1 = ( 1 , 0 , … , 0 , 0 ) (del(x^(1),dots,x^(n)))/(delx^(1))=(1,0,dots,0,0)\frac{\partial\left(x^{1}, \ldots, x^{n}\right)}{\partial x^{1}}=(1,0, \ldots, 0,0)∂(x1,…,xn)∂x1=(1,0,…,0,0)
Similarly, the tangent vector that takes x 2 x 2 x^(2)x^{2}x2 to 1 and all the other coordinates to 0 is the coordinate basis vector e 2 = ∂ ∂ x 2 e 2 = ∂ ∂ x 2 e_(2)=(del)/(delx^(2))\mathbf{e}_{2}=\frac{\partial}{\partial x^{2}}e2=∂∂x2 and has components
∂ ( x 1 , … , x n ) ∂ x 2 = ( 0 , 1 , … , 0 , 0 ) ∂ x 1 , … , x n ∂ x 2 = ( 0 , 1 , … , 0 , 0 ) (del(x^(1),dots,x^(n)))/(delx^(2))=(0,1,dots,0,0)\frac{\partial\left(x^{1}, \ldots, x^{n}\right)}{\partial x^{2}}=(0,1, \ldots, 0,0)∂(x1,…,xn)∂x2=(0,1,…,0,0)
And the tangent vector that takes x n x n x^(n)x^{n}xn to 1 and all the other coordinates to 0 is the coordinate basis vector e n = ∂ ∂ x n e n = ∂ ∂ x n e_(n)=(del)/(delx^(n))\mathbf{e}_{n}=\frac{\partial}{\partial x^{n}}en=∂∂xn and has components
∂ ( x 1 , … , x n ) ∂ x n = ( 0 , 0 , … , 0 , 1 ) ∂ x 1 , … , x n ∂ x n = ( 0 , 0 , … , 0 , 1 ) (del(x^(1),dots,x^(n)))/(delx^(n))=(0,0,dots,0,1)\frac{\partial\left(x^{1}, \ldots, x^{n}\right)}{\partial x^{n}}=(0,0, \ldots, 0,1)∂(x1,…,xn)∂xn=(0,0,…,0,1)
In other words, the coordinate basis vectors e i = ∂ ∂ x i e i = ∂ ∂ x i e_(i)=(del)/(delx^(i))\mathbf{e}_{i}=\frac{\partial}{\partial x^{i}}ei=∂∂xi are represented by the standard basis vectors (1.2.1) of R n : R n : R^(n):\mathbb{R}^{n}:Rn:
e 1 = ( 1 , 0 , … , 0 , 0 ) , e 2 = ( 0 , 1 , … , 0 , 0 ) , … , e n = ( 0 , 0 , … , 0 , 1 ) e 1 = ( 1 , 0 , … , 0 , 0 ) , e 2 = ( 0 , 1 , … , 0 , 0 ) , … , e n = ( 0 , 0 , … , 0 , 1 ) e_(1)=(1,0,dots,0,0),e_(2)=(0,1,dots,0,0),dots,e_(n)=(0,0,dots,0,1)\mathbf{e}_{1}=(1,0, \ldots, 0,0), \mathbf{e}_{2}=(0,1, \ldots, 0,0), \ldots, \mathbf{e}_{n}=(0,0, \ldots, 0,1)e1=(1,0,…,0,0),e2=(0,1,…,0,0),…,en=(0,0,…,0,1)
Finally, it's worth mentioning that writing basis vectors as ∂ ∂ x i ∂ ∂ x i (del)/(delx^(i))\frac{\partial}{\partial x^{i}}∂∂xi means we can easily derive the change of basis formula ( 2.2 .6 ) ( 2.2 .6 ) (2.2.6)(2.2 .6)(2.2.6)
( e r , e θ ) = ( e ^ x , e ^ y ) [ ∂ x ∂ r ∂ x ∂ θ ∂ y ∂ r ∂ y ∂ θ ] e r , e θ = e ^ x , e ^ y ∂ x ∂ r      ∂ x ∂ θ ∂ y ∂ r      ∂ y ∂ θ (e_(r),e_(theta))=( hat(e)_(x), hat(e)_(y))[[(del x)/(del r),(del x)/(del theta)],[(del y)/(del r),(del y)/(del theta)]]\left(\mathbf{e}_{r}, \mathbf{e}_{\theta}\right)=\left(\hat{\mathbf{e}}_{x}, \hat{\mathbf{e}}_{y}\right)\left[\begin{array}{ll} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{array}\right](er,eθ)=(e^x,e^y)[∂x∂r∂x∂θ∂y∂r∂y∂θ]
we used in section 2.2.4. Instead of writing
e r = ∂ x ∂ r e ^ x + ∂ y ∂ r e ^ y e r = ∂ x ∂ r e ^ x + ∂ y ∂ r e ^ y e_(r)=(del x)/(del r) hat(e)_(x)+(del y)/(del r) hat(e)_(y)\mathbf{e}_{r}=\frac{\partial x}{\partial r} \hat{\mathbf{e}}_{x}+\frac{\partial y}{\partial r} \hat{\mathbf{e}}_{y}er=∂x∂re^x+∂y∂re^y
we write
∂ ∂ r = ∂ x ∂ r ∂ ∂ x + ∂ y ∂ r ∂ ∂ y ∂ ∂ r = ∂ x ∂ r ∂ ∂ x + ∂ y ∂ r ∂ ∂ y (del)/(del r)=(del x)/(del r)(del)/(del x)+(del y)/(del r)(del)/(del y)\frac{\partial}{\partial r}=\frac{\partial x}{\partial r} \frac{\partial}{\partial x}+\frac{\partial y}{\partial r} \frac{\partial}{\partial y}∂∂r=∂x∂r∂∂x+∂y∂r∂∂y
which is a version of the partial derivatives chain rule. Similarly, for
e θ = ∂ x ∂ θ e ^ x + ∂ y ∂ θ e ^ y e θ = ∂ x ∂ θ e ^ x + ∂ y ∂ θ e ^ y e_(theta)=(del x)/(del theta) hat(e)_(x)+(del y)/(del theta) hat(e)_(y)\mathbf{e}_{\theta}=\frac{\partial x}{\partial \theta} \hat{\mathbf{e}}_{x}+\frac{\partial y}{\partial \theta} \hat{\mathbf{e}}_{y}eθ=∂x∂θe^x+∂y∂θe^y
we write
∂ ∂ θ = ∂ x ∂ θ ∂ ∂ x + ∂ y ∂ θ ∂ ∂ y ∂ ∂ θ = ∂ x ∂ θ ∂ ∂ x + ∂ y ∂ θ ∂ ∂ y (del)/(del theta)=(del x)/(del theta)(del)/(del x)+(del y)/(del theta)(del)/(del y)\frac{\partial}{\partial \theta}=\frac{\partial x}{\partial \theta} \frac{\partial}{\partial x}+\frac{\partial y}{\partial \theta} \frac{\partial}{\partial y}∂∂θ=∂x∂θ∂∂x+∂y∂θ∂∂y
Using index notation, the general change of basis formula, where ( x 1 , … , x n ) x 1 , … , x n (x^(1),dots,x^(n))\left(x^{1}, \ldots, x^{n}\right)(x1,…,xn) and ( y 1 , … , y n ) y 1 , … , y n (y^(1),dots,y^(n))\left(y^{1}, \ldots, y^{n}\right)(y1,…,yn) are, respectively, the old and new coordinate systems can be succinctly written 1 1 ^(1)^{1}1 as
∂ ∂ y j = ∂ x i ∂ y j ∂ ∂ x i ∂ ∂ y j = ∂ x i ∂ y j ∂ ∂ x i (del)/(dely^(j))=(delx^(i))/(dely^(j))(del)/(delx^(i))\frac{\partial}{\partial y^{j}}=\frac{\partial x^{i}}{\partial y^{j}} \frac{\partial}{\partial x^{i}}∂∂yj=∂xi∂yj∂∂xi which is equivalent to the matrix equation
( ∂ ∂ y 1 , … , ∂ ∂ y n ) = ( ∂ ∂ x 1 , … , ∂ ∂ x n ) [ ∂ x 1 ∂ y 1 ⋯ ∂ x 1 ∂ y n ⋮ ⋮ ∂ x n ∂ y 1 ⋯ ∂ x n ∂ y n ] ∂ ∂ y 1 , … , ∂ ∂ y n = ∂ ∂ x 1 , … , ∂ ∂ x n ∂ x 1 ∂ y 1 ⋯ ∂ x 1 ∂ y n ⋮ ⋮ ∂ x n ∂ y 1 ⋯ ∂ x n ∂ y n ((del)/(dely^(1)),dots,(del)/(dely^(n)))=((del)/(delx^(1)),dots,(del)/(delx^(n)))[[(delx^(1))/(dely^(1)),cdots,(delx^(1))/(dely^(n))],[vdots,,vdots],[(delx^(n))/(dely^(1)),cdots,(delx^(n))/(dely^(n))]]\left(\frac{\partial}{\partial y^{1}}, \ldots, \frac{\partial}{\partial y^{n}}\right)=\left(\frac{\partial}{\partial x^{1}}, \ldots, \frac{\partial}{\partial x^{n}}\right)\left[\begin{array}{ccc} \frac{\partial x^{1}}{\partial y^{1}} & \cdots & \frac{\partial x^{1}}{\partial y^{n}} \\ \vdots & & \vdots \\ \frac{\partial x^{n}}{\partial y^{1}} & \cdots & \frac{\partial x^{n}}{\partial y^{n}} \end{array}\right](∂∂y1,…,∂∂yn)=(∂∂x1,…,∂∂xn)[∂x1∂y1⋯∂x1∂yn⋮⋮∂xn∂y1⋯∂xn∂yn]
where the matrix of partial derivatives
∂ x i ∂ y j = [ ∂ x 1 ∂ y 1 ⋯ ∂ x 1 ∂ y n ⋮ ⋮ ∂ x n ∂ y 1 ⋯ ∂ x n ∂ y n ] ∂ x i ∂ y j = ∂ x 1 ∂ y 1 ⋯ ∂ x 1 ∂ y n ⋮ ⋮ ∂ x n ∂ y 1 ⋯ ∂ x n ∂ y n (delx^(i))/(dely^(j))=[[(delx^(1))/(dely^(1)),cdots,(delx^(1))/(dely^(n))],[vdots,,vdots],[(delx^(n))/(dely^(1)),cdots,(delx^(n))/(dely^(n))]]\frac{\partial x^{i}}{\partial y^{j}}=\left[\begin{array}{ccc} \frac{\partial x^{1}}{\partial y^{1}} & \cdots & \frac{\partial x^{1}}{\partial y^{n}} \\ \vdots & & \vdots \\ \frac{\partial x^{n}}{\partial y^{1}} & \cdots & \frac{\partial x^{n}}{\partial y^{n}} \end{array}\right]∂xi∂yj=[∂x1∂y1⋯∂x1∂yn⋮⋮∂xn∂y1⋯∂xn∂yn]
is called the Jacobian matrix.
Even though we don't use the ∂ ∂ x i ∂ ∂ x i (del)/(delx^(i))\frac{\partial}{\partial x^{i}}∂∂xi notation, we are nevertheless making implicit use of coordinate basis vectors throughout this book. Coordinates basis vectors are widely used in more advanced mathematics and physics (relativity theory, for example).

2.2.7 Another 1 -form example

Figure 2.4: Work form ω ω omega\omegaω of force field F F F\mathbf{F}F.
If F F F\mathbf{F}F is a force field in R 3 R 3 R^(3)\mathbb{R}^{3}R3 and v v v\mathbf{v}v is a displacement vector, the work corresponding to this displacement is given by the dot product F ⋅ v F ⋅ v F*v\mathbf{F} \cdot \mathbf{v}F⋅v (see Figure 2.4). We can also describe the work in terms of a 1 -form ω ω omega\omegaω acting on v v v\mathbf{v}v, where
ω ( v ) = F ⋅ v ω ( v ) = F ⋅ v omega(v)=F*v\omega(\mathbf{v})=\mathbf{F} \cdot \mathbf{v}ω(v)=F⋅v
So if we feed vector v v v\mathbf{v}v to the 1 -form ω ω omega\omegaω, out will pop a number - the work corresponding to the displacement v v v\mathbf{v}v. The 1 -form ω ω omega\omegaω is called the work form or force 1-form of the force field F F F\mathbf{F}F.
In R 3 R 3 R^(3)\mathbb{R}^{3}R3 all 1-forms are work forms of some vector field.

2.3 2-forms

2-forms and higher degree forms are multilinear alternating functions of tangent vectors. So what does that mean in plain English? Take a 3-form ω ω omega\omegaω, for example, feed it three vectors u , v u , v u,v\mathbf{u}, \mathbf{v}u,v and w w w\mathbf{w}w, and out pops a number. For the sake of argument, let's assume that number is 7 , ie ω ( u , v , w ) = 7 ω ( u , v , w ) = 7 omega(u,v,w)=7\omega(\mathbf{u}, \mathbf{v}, \mathbf{w})=7ω(u,v,w)=7.
  • Alternating (or antisymmetric or skew-symmetric) means if we exchange any two of the vectors, the number changes sign. For example, ω ( v , u , w ) = − 7 ω ( v , u , w ) = − 7 omega(v,u,w)=-7\omega(\mathbf{v}, \mathbf{u}, \mathbf{w})=-7ω(v,u,w)=−7.
  • Multilinear means that if we feed the same 3 -form the vectors 2 u , v 2 u , v 2u,v2 \mathbf{u}, \mathbf{v}2u,v and w w w\mathbf{w}w, the number changes to 2 × 7 = 14 2 × 7 = 14 2xx7=142 \times 7=142×7=14; feed ω ω omega\omegaω the vectors 3 u , 2 v 3 u , 2 v 3u,2v3 \mathbf{u}, 2 \mathbf{v}3u,2v and 5 w 5 w 5w5 \mathbf{w}5w and the number is 3 × 2 × 5 × 7 = 210 3 × 2 × 5 × 7 = 210 3xx2xx5xx7=2103 \times 2 \times 5 \times 7=2103×2×5×7=210.
A 2 -form ω ω omega\omegaω is:
  • a function on a pair of vectors, which is bilinear (ie multilinear for two variables) and alternating, ie
ω ( a u + b v , w ) = a ω ( u , w ) + b ω ( v , w ) ω ( u , v ) = − ω ( v , u ) ω ( a u + b v , w ) = a ω ( u , w ) + b ω ( v , w ) ω ( u , v ) = − ω ( v , u ) {:[omega(au+bv","w)=a omega(u","w)+b omega(v","w)],[omega(u","v)=-omega(v","u)]:}\begin{gathered} \omega(a \mathbf{u}+b \mathbf{v}, \mathbf{w})=a \omega(\mathbf{u}, \mathbf{w})+b \omega(\mathbf{v}, \mathbf{w}) \\ \omega(\mathbf{u}, \mathbf{v})=-\omega(\mathbf{v}, \mathbf{u}) \end{gathered}ω(au+bv,w)=aω(u,w)+bω(v,w)ω(u,v)=−ω(v,u)
where a a aaa and b b bbb are real numbers.
On R 3 R 3 R^(3)\mathbb{R}^{3}R3 a 2-form will include one or more of the basis 2-forms d y ∧ d z , d z ∧ d x d y ∧ d z , d z ∧ d x dy^^dz,dz^^dxd y \wedge d z, d z \wedge d xdy∧dz,dz∧dx and d x ∧ d y d x ∧ d y dx^^dyd x \wedge d ydx∧dy and look something like
ω = f 1 d y ∧ d z + f 2 d z ∧ d x + f 3 d x ∧ d y ω = f 1 d y ∧ d z + f 2 d z ∧ d x + f 3 d x ∧ d y omega=f_(1)dy^^dz+f_(2)dz^^dx+f_(3)dx^^dy\omega=f_{1} d y \wedge d z+f_{2} d z \wedge d x+f_{3} d x \wedge d yω=f1dy∧dz+f2dz∧dx+f3dx∧dy
where the ∧ ∧ ^^\wedge∧ symbol denotes a type of multiplication called the wedge product. As we'll see shortly, the wedge product is anti-commutative (ie d x ∧ d y = − d y ∧ d x d x ∧ d y = − d y ∧ d x dx^^dy=-dy^^dxd x \wedge d y=-d y \wedge d xdx∧dy=−dy∧dx ), meaning the order of the differentials is not arbitrary.
So
ω = 5 d y ∧ d z ω = 5 x y d z ∧ d x − z d x ∧ d y ω = 5 d y ∧ d z ω = 5 x y d z ∧ d x − z d x ∧ d y {:[omega=5dy^^dz],[omega=5xydz^^dx-zdx^^dy]:}\begin{gathered} \omega=5 d y \wedge d z \\ \omega=5 x y d z \wedge d x-z d x \wedge d y \end{gathered}ω=5dy∧dzω=5xydz∧dx−zdx∧dy
and
ω = 2 d y ∧ d z + 3 y d z ∧ d x + d x ∧ d y ω = 2 d y ∧ d z + 3 y d z ∧ d x + d x ∧ d y omega=2dy^^dz+3ydz^^dx+dx^^dy\omega=2 d y \wedge d z+3 y d z \wedge d x+d x \wedge d yω=2dy∧dz+3ydz∧dx+dx∧dy
are all examples of 2 -forms.
We can create a 2 -form by multiplying (using the wedge product) two 1 -forms together. For example,
6 x d x ∧ 3 d y = 18 x d x ∧ d y 6 x d x ∧ 3 d y = 18 x d x ∧ d y 6xdx^^3dy=18 xdx^^dy6 x d x \wedge 3 d y=18 x d x \wedge d y6xdx∧3dy=18xdx∧dy
In this integral
∫ S 3 x y d x ∧ d y ⏟ 2 − form ∫ S   3 x y d x ∧ d y ⏟ 2 −  form  int_(S)ubrace(3xydx^^dyubrace)_(2-" form ")\int_{S} \underbrace{3 x y d x \wedge d y}_{2-\text { form }}∫S3xydx∧dy⏟2− form 
3 x y d x ∧ d y 3 x y d x ∧ d y 3xydx^^dy3 x y d x \wedge d y3xydx∧dy is a 2 -form (recall that S S SSS denotes a two-dimensional surface in R n R n R^(n)\mathbb{R}^{n}Rn ).

2.3.1 The wedge product

We need a way of multiplying a p p ppp-form with a q q qqq-form to give a ( p + q ) ( p + q ) (p+q)(p+q)(p+q)-form. For example, we need to be able to multiply two 1-forms to give a 2 -form. We do this using a type of multiplication known as the wedge product or exterior product, denoted by the symbol ∧ ∧ ^^\wedge∧.
There's bad news and good news regarding the wedge product. The bad news is that the formula for the wedge product is tricky. Hubbard and Hubbard [11] state:
The wedge product is a messy thing: a complicated summation, over various shuffles of vectors, of the product of two k k kkk-forms ...
However, the good news is that we don't need to worry about the formula for the wedge product. The essential takeaway for us is that the wedge product is anti-commutative for the differentials d x , d y , d z d x , d y , d z dx,dy,dzd x, d y, d zdx,dy,dz, etc. So, for example,
d x ∧ d y = − d y ∧ d x d x ∧ d y = − d y ∧ d x dx^^dy=-dy^^dxd x \wedge d y=-d y \wedge d xdx∧dy=−dy∧dx
Which means, in turn, that the wedge product is anti-commutative for the general 1 -forms ω ω omega\omegaω and ν ν nu\nuν, ie
ω ∧ ν = − ν ∧ ω ω ∧ ν = − ν ∧ ω omega^^nu=-nu^^omega\omega \wedge \nu=-\nu \wedge \omegaω∧ν=−ν∧ω
Algebra that makes use of the wedge product is known as exterior algebra. Thankfully, the only difference between exterior and ordinary algebra is the anti-commutativity rule.
The intuitive motivation for the wedge product is that it allows n n nnn vector-like objects (vectors or differential forms, for example) to be combined in a geometrical fashion to give an n n nnn-dimensional area or volume. The wedge product of two vectors u u u\mathbf{u}u and v v v\mathbf{v}v can be represented by a parallelogram spanned by u u u\mathbf{u}u and v v v\mathbf{v}v as shown in Figure 2.5
Notice that the order of the two vectors matters: u ∧ v u ∧ v u^^v\mathbf{u} \wedge \mathbf{v}u∧v has a clockwise orientation, v ∧ u v ∧ u v^^u\mathbf{v} \wedge \mathbf{u}v∧u has a counterclockwise orientation. The notion of orientation, which we'll discuss in chapter 6 , is the rationale for the wedge product anti-commutativity rule.
We can informally derive that rule by considering d x d x dxd xdx as an infinitesimal displacement along the x x xxx axis. The parallelogram spanned by two such d x d x dxd xdx will have zero area (a
Figure 2.5: The wedge product and orientation.
parallelogram with sides spanned by the same vector is a straight line). We can write this as
d x ∧ d x = 0 d x ∧ d x = 0 dx^^dx=0d x \wedge d x=0dx∧dx=0
It must follow that
( d x + d y ) ∧ ( d x + d y ) = 0 ( d x + d y ) ∧ ( d x + d y ) = 0 (dx+dy)^^(dx+dy)=0(d x+d y) \wedge(d x+d y)=0(dx+dy)∧(dx+dy)=0
which we can expand to get
( d x ∧ d x ) + ( d x ∧ d y ) + ( d y ∧ d x ) + ( d y ∧ d y ) = 0 ( d x ∧ d x ) + ( d x ∧ d y ) + ( d y ∧ d x ) + ( d y ∧ d y ) = 0 (dx^^dx)+(dx^^dy)+(dy^^dx)+(dy^^dy)=0(d x \wedge d x)+(d x \wedge d y)+(d y \wedge d x)+(d y \wedge d y)=0(dx∧dx)+(dx∧dy)+(dy∧dx)+(dy∧dy)=0
Because
( d x ∧ d y ) + ( d y ∧ d x ) = 0 ( d x ∧ d y ) + ( d y ∧ d x ) = 0 (dx^^dy)+(dy^^dx)=0(d x \wedge d y)+(d y \wedge d x)=0(dx∧dy)+(dy∧dx)=0
we have
d x ∧ d y = − d y ∧ d x d x ∧ d y = − d y ∧ d x dx^^dy=-dy^^dxd x \wedge d y=-d y \wedge d xdx∧dy=−dy∧dx
and we have shown that the wedge product is anti-commutative.
Using index notation, we can generalise d x ∧ d x = 0 d x ∧ d x = 0 dx^^dx=0d x \wedge d x=0dx∧dx=0 to
(https://cdn.mathpix.com/cropped/2024_05_19_038e0d3438e9b54bb93dg-39.jpg?height=43&width=95&top_left_y=1450&top_left_x=1452) d x i ∧ d x i = 0 (https://cdn.mathpix.com/cropped/2024_05_19_038e0d3438e9b54bb93dg-39.jpg?height=43&width=95&top_left_y=1450&top_left_x=1452) d x i ∧ d x i = 0 {:(https://cdn.mathpix.com/cropped/2024_05_19_038e0d3438e9b54bb93dg-39.jpg?height=43&width=95&top_left_y=1450&top_left_x=1452)dx^(i)^^dx^(i)=0:}\begin{equation*} d x^{i} \wedge d x^{i}=0 \tag{https://cdn.mathpix.com/cropped/2024_05_19_038e0d3438e9b54bb93dg-39.jpg?height=43&width=95&top_left_y=1450&top_left_x=1452} \end{equation*}(https://cdn.mathpix.com/cropped/2024_05_19_038e0d3438e9b54bb93dg-39.jpg?height=43&width=95&top_left_y=1450&top_left_x=1452)dxi∧dxi=0
and conclude that the wedge product of any 1 -form with itself is zero.
Functions commute with d x d x dxd xdx, ie
f ∧ d x = d x ∧ f f ∧ d x = d x ∧ f f^^dx=dx^^ff \wedge d x=d x \wedge ff∧dx=dx∧f
The convention is not to use the ∧ ∧ ^^\wedge∧ when multiplying a form by a function, ie to write f d x f d x fdxf d xfdx and not f ∧ d x f ∧ d x f^^dxf \wedge d xf∧dx.
Example 2.1. Calculate the wedge product ω ∧ ν ω ∧ ν omega^^nu\omega \wedge \nuω∧ν and ν ∧ ω ν ∧ ω nu^^omega\nu \wedge \omegaν∧ω of the 1-forms ω = 3 d x − 5 d y ω = 3 d x − 5 d y omega=3dx-5dy\omega=3 d x-5 d yω=3dx−5dy and ν = d x + 2 d y ν = d x + 2 d y nu=dx+2dy\nu=d x+2 d yν=dx+2dy.
ω ∧ ν = ( 3 d x − 5 d y ) ( d x + 2 d y ) = 3 d x ∧ d x + 6 d x ∧ d y − 5 d y ∧ d x − 10 d y ∧ d y ω ∧ ν = ( 3 d x − 5 d y ) ( d x + 2 d y ) = 3 d x ∧ d x + 6 d x ∧ d y − 5 d y ∧ d x − 10 d y ∧ d y {:[omega^^nu=(3dx-5dy)(dx+2dy)],[=3dx^^dx+6dx^^dy-5dy^^dx-10 dy^^dy]:}\begin{gathered} \omega \wedge \nu=(3 d x-5 d y)(d x+2 d y) \\ =3 d x \wedge d x+6 d x \wedge d y-5 d y \wedge d x-10 d y \wedge d y \end{gathered}ω∧ν=(3dx−5dy)(dx+2dy)=3dx∧dx+6dx∧dy−5dy∧dx−10dy∧dy
= 11 d x ∧ d y = 11 d x ∧ d y =11 dx^^dy=11 d x \wedge d y=11dx∧dy
because d x ∧ d x = d y ∧ d y = 0 d x ∧ d x = d y ∧ d y = 0 dx^^dx=dy^^dy=0d x \wedge d x=d y \wedge d y=0dx∧dx=dy∧dy=0 and − 5 d y ∧ d x = 5 d x ∧ d y − 5 d y ∧ d x = 5 d x ∧ d y -5dy^^dx=5dx^^dy-5 d y \wedge d x=5 d x \wedge d y−5dy∧dx=5dx∧dy.
ν ∧ ω = ( d x + 2 d y ) ( 3 d x − 5 d y ) = 3 d x ∧ d x − 5 d x ∧ d y + 6 d y ∧ d x − 10 d y ∧ d y = − 11 d x ∧ d y ν ∧ ω = ( d x + 2 d y ) ( 3 d x − 5 d y ) = 3 d x ∧ d x − 5 d x ∧ d y + 6 d y ∧ d x − 10 d y ∧ d y = − 11 d x ∧ d y {:[nu^^omega=(dx+2dy)(3dx-5dy)],[=3dx^^dx-5dx^^dy+6dy^^dx-10 dy^^dy],[=-11 dx^^dy]:}\begin{gathered} \nu \wedge \omega=(d x+2 d y)(3 d x-5 d y) \\ =3 d x \wedge d x-5 d x \wedge d y+6 d y \wedge d x-10 d y \wedge d y \\ =-11 d x \wedge d y \end{gathered}ν∧ω=(dx+2dy)(3dx−5dy)=3dx∧dx−5dx∧dy+6dy∧dx−10dy∧dy=−11dx∧dy
because d x ∧ d x = d y ∧ d y = 0 d x ∧ d x = d y ∧ d y = 0 dx^^dx=dy^^dy=0d x \wedge d x=d y \wedge d y=0dx∧dx=dy∧dy=0 and 6 d y ∧ d x = − 6 d x ∧ d y 6 d y ∧ d x = − 6 d x ∧ d y 6dy^^dx=-6dx^^dy6 d y \wedge d x=-6 d x \wedge d y6dy∧dx=−6dx∧dy. And we can see that ω ∧ ν = − ν ∧ ω ω ∧ ν = − ν ∧ ω omega^^nu=-nu^^omega\omega \wedge \nu=-\nu \wedge \omegaω∧ν=−ν∧ω.
Example 2.2. Calculate the wedge product of the 1-forms ω = u 1 d x + u 2 d y + u 3 d z ω = u 1 d x + u 2 d y + u 3 d z omega=u_(1)dx+u_(2)dy+u_(3)dz\omega=u_{1} d x+u_{2} d y+u_{3} d zω=u1dx+u2dy+u3dz and ν = v 1 d x + v 2 d y + v 3 d z ν = v 1 d x + v 2 d y + v 3 d z nu=v_(1)dx+v_(2)dy+v_(3)dz\nu=v_{1} d x+v_{2} d y+v_{3} d zν=v1dx+v2dy+v3dz.
ω ∧ ν = ( u 1 d x + u 2 d y + u 3 d z ) ∧ ( v 1 d x + v 2 d y + v 3 d z ) = u 1 d x ∧ v 1 d x + u 1 d x ∧ v 2 d y + u 1 d x ∧ v 3 d z + u 2 d y ∧ v 1 d x + u 2 d y ∧ v 2 d y + u 2 d y ∧ v 3 d z + u 3 d z ∧ v 1 d x + u 3 d z ∧ v 2 d y + u 3 d z ∧ v 3 d z ω ∧ ν = u 1 d x + u 2 d y + u 3 d z ∧ v 1 d x + v 2 d y + v 3 d z = u 1 d x ∧ v 1 d x + u 1 d x ∧ v 2 d y + u 1 d x ∧ v 3 d z + u 2 d y ∧ v 1 d x + u 2 d y ∧ v 2 d y + u 2 d y ∧ v 3 d z + u 3 d z ∧ v 1 d x + u 3 d z ∧ v 2 d y + u 3 d z ∧ v 3 d z {:[omega^^nu=(u_(1)dx+u_(2)dy+u_(3)dz)^^(v_(1)dx+v_(2)dy+v_(3)dz)],[=u_(1)dx^^v_(1)dx+u_(1)dx^^v_(2)dy+u_(1)dx^^v_(3)dz],[+u_(2)dy^^v_(1)dx+u_(2)dy^^v_(2)dy+u_(2)dy^^v_(3)dz],[+u_(3)dz^^v_(1)dx+u_(3)dz^^v_(2)dy+u_(3)dz^^v_(3)dz]:}\begin{aligned} & \omega \wedge \nu=\left(u_{1} d x+u_{2} d y+u_{3} d z\right) \wedge\left(v_{1} d x+v_{2} d y+v_{3} d z\right) \\ &=u_{1} d x \wedge v_{1} d x+u_{1} d x \wedge v_{2} d y+u_{1} d x \wedge v_{3} d z \\ &+u_{2} d y \wedge v_{1} d x+u_{2} d y \wedge v_{2} d y+u_{2} d y \wedge v_{3} d z \\ &+u_{3} d z \wedge v_{1} d x+u_{3} d z \wedge v_{2} d y+u_{3} d z \wedge v_{3} d z \end{aligned}ω∧ν=(u1dx+u2dy+u3dz)∧(v1dx+v2dy+v3dz)=u1dx∧v1dx+u1dx∧v2dy+u1dx∧v3dz+u2dy∧v1dx+u2dy∧v2dy+u2dy∧v3dz+u3dz∧v1dx+u3dz∧v2dy+u3dz∧v3dz
Which simplifies to give the 2-form
ω ∧ ν = ( u 2 v 3 − u 3 v 2 ) d y ∧ d z + ( u 3 v 1 − u 1 v 3 ) d z ∧ d x + ( u 1 v 2 − u 2 v 1 ) d x ∧ d y ω ∧ ν = u 2 v 3 − u 3 v 2 d y ∧ d z + u 3 v 1 − u 1 v 3 d z ∧ d x + u 1 v 2 − u 2 v 1 d x ∧ d y omega^^nu=(u_(2)v_(3)-u_(3)v_(2))dy^^dz+(u_(3)v_(1)-u_(1)v_(3))dz^^dx+(u_(1)v_(2)-u_(2)v_(1))dx^^dy\omega \wedge \nu=\left(u_{2} v_{3}-u_{3} v_{2}\right) d y \wedge d z+\left(u_{3} v_{1}-u_{1} v_{3}\right) d z \wedge d x+\left(u_{1} v_{2}-u_{2} v_{1}\right) d x \wedge d yω∧ν=(u2v3−u3v2)dy∧dz+(u3v1−u1v3)dz∧dx+(u1v2−u2v1)dx∧dy
because u 1 d x ∧ v 1 d x = u 2 d y ∧ v 2 d y = u 3 d z ∧ v 3 d z = 0 u 1 d x ∧ v 1 d x = u 2 d y ∧ v 2 d y = u 3 d z ∧ v 3 d z = 0 u_(1)dx^^v_(1)dx=u_(2)dy^^v_(2)dy=u_(3)dz^^v_(3)dz=0u_{1} d x \wedge v_{1} d x=u_{2} d y \wedge v_{2} d y=u_{3} d z \wedge v_{3} d z=0u1dx∧v1dx=u2dy∧v2dy=u3dz∧v3dz=0 and d y ∧ d z = − d z ∧ d y d y ∧ d z = − d z ∧ d y dy^^dz=-dz^^dyd y \wedge d z=-d z \wedge d ydy∧dz=−dz∧dy, d z ∧ d x = − d x ∧ d z d z ∧ d x = − d x ∧ d z dz^^dx=-dx^^dzd z \wedge d x=-d x \wedge d zdz∧dx=−dx∧dz and d x ∧ d y = − d y ∧ d x d x ∧ d y = − d y ∧ d x dx^^dy=-dy^^dxd x \wedge d y=-d y \wedge d xdx∧dy=−dy∧dx.
Notice that if we let u 1 = u 1 , u 2 = u 2 , u 3 = u 3 , v 1 = v 1 , v 2 = v 2 u 1 = u 1 , u 2 = u 2 , u 3 = u 3 , v 1 = v 1 , v 2 = v 2 u_(1)=u^(1),u_(2)=u^(2),u_(3)=u^(3),v_(1)=v^(1),v_(2)=v^(2)u_{1}=u^{1}, u_{2}=u^{2}, u_{3}=u^{3}, v_{1}=v^{1}, v_{2}=v^{2}u1=u1,u2=u2,u3=u3,v1=v1,v2=v2 and v 3 = v 3 v 3 = v 3 v_(3)=v^(3)v_{3}=v^{3}v3=v3, then the coefficients of this 2 -form ω ∧ ν ω ∧ ν omega^^nu\omega \wedge \nuω∧ν are the same as the components of the cross product (1.6.1) of the vectors u = u 1 e ^ x + u 2 e ^ y + u 3 e ^ z u = u 1 e ^ x + u 2 e ^ y + u 3 e ^ z u=u^(1) hat(e)_(x)+u^(2) hat(e)_(y)+u^(3) hat(e)_(z)\mathbf{u}=u^{1} \hat{\mathbf{e}}_{x}+u^{2} \hat{\mathbf{e}}_{y}+u^{3} \hat{\mathbf{e}}_{z}u=u1e^x+u2e^y+u3e^z and v = v 1 e ^ x + v 2 e ^ y + v 3 e ^ z v = v 1 e ^ x + v 2 e ^ y + v 3 e ^ z v=v^(1) hat(e)_(x)+v^(2) hat(e)_(y)+v^(3) hat(e)_(z)\mathbf{v}=v^{1} \hat{\mathbf{e}}_{x}+v^{2} \hat{\mathbf{e}}_{y}+v^{3} \hat{\mathbf{e}}_{z}v=v1e^x+v2e^y+v3e^z :
u × v = ( u 2 v 3 − u 3 v 2 ) e ^ x + ( u 3 v 1 − u 1 v 3 ) e ^ y + ( u 1 v 2 − u 2 v 1 ) e ^ z u × v = u 2 v 3 − u 3 v 2 e ^ x + u 3 v 1 − u 1 v 3 e ^ y + u 1 v 2 − u 2 v 1 e ^ z uxxv=(u^(2)v^(3)-u^(3)v^(2)) hat(e)_(x)+(u^(3)v^(1)-u^(1)v^(3)) hat(e)_(y)+(u^(1)v^(2)-u^(2)v^(1)) hat(e)_(z)\mathbf{u} \times \mathbf{v}=\left(u^{2} v^{3}-u^{3} v^{2}\right) \hat{\mathbf{e}}_{x}+\left(u^{3} v^{1}-u^{1} v^{3}\right) \hat{\mathbf{e}}_{y}+\left(u^{1} v^{2}-u^{2} v^{1}\right) \hat{\mathbf{e}}_{z}u×v=(u2v3−u3v2)e^x+(u3v1−u1v3)e^y+(u1v2−u2v1)e^z
We can therefore regard the wedge product as an n n nnn-dimensional generalisation of the three-dimensional cross product.
Moving on, we can now see why there are no 4 -forms or higher in R 3 R 3 R^(3)\mathbb{R}^{3}R3. If we multiply together any four of the differentials d x , d y , d z d x , d y , d z dx,dy,dzd x, d y, d zdx,dy,dz, we must have at least two of the same and the result will be zero. For example,
d x ∧ d y ∧ d z ∧ d x = − d x ∧ d y ∧ d x ∧ d z = d x ∧ d x ∧ d y ∧ d z = 0 d x ∧ d y ∧ d z ∧ d x = − d x ∧ d y ∧ d x ∧ d z = d x ∧ d x ∧ d y ∧ d z = 0 {:[dx^^dy^^dz^^dx],[=-dx^^dy^^dx^^dz],[=dx^^dx^^dy^^dz=0]:}\begin{gathered} d x \wedge d y \wedge d z \wedge d x \\ =-d x \wedge d y \wedge d x \wedge d z \\ =d x \wedge d x \wedge d y \wedge d z=0 \end{gathered}dx∧dy∧dz∧dx=−dx∧dy∧dx∧dz=dx∧dx∧dy∧dz=0
because d x ∧ d x = 0 d x ∧ d x = 0 dx^^dx=0d x \wedge d x=0dx∧dx=0.
Therefore, an n n nnn-dimensional space M M MMM can have up to n n nnn-dimensional differential forms.
Figure 2.6: d x ∧ d y d x ∧ d y dx^^dyd x \wedge d ydx∧dy acting on vectors u u u\mathbf{u}u and v v v\mathbf{v}v in R 2 R 2 R^(2)\mathbb{R}^{2}R2.

2.3.2 2-forms acting on two vectors

We've seen how a 1-form acts on a vector to give a number. A 2 -form acts on a pair of vectors to give a number, ie a 2 -form is a function that requires two vector arguments. Let's see how this works in R 2 R 2 R^(2)\mathbb{R}^{2}R2 with a couple of vectors u = ( u 1 , u 2 ) u = u 1 , u 2 u=(u^(1),u^(2))\mathbf{u}=\left(u^{1}, u^{2}\right)u=(u1,u2) and v = ( v 1 , v 2 ) v = v 1 , v 2 v=(v^(1),v^(2))\mathbf{v}=\left(v^{1}, v^{2}\right)v=(v1,v2), as shown in Figure 2.6. A natural number defined from u u u\mathbf{u}u and v v v\mathbf{v}v is the (signed) area of the parallelogram they span. We know from when we looked at determinants in section 1.7 that the signed area of the parallelogram spanned by u u u\mathbf{u}u and v v v\mathbf{v}v is given by
| u 1 v 1 u 2 v 2 | = u 1 v 2 − u 2 v 1 u 1      v 1 u 2      v 2 = u 1 v 2 − u 2 v 1 |[u^(1),v^(1)],[u^(2),v^(2)]|=u^(1)v^(2)-u^(2)v^(1)\left|\begin{array}{ll} u^{1} & v^{1} \\ u^{2} & v^{2} \end{array}\right|=u^{1} v^{2}-u^{2} v^{1}|u1v1u2v2|=u1v2−u2v1
Because d x i d x i dx^(i)d x^{i}dxi picks out a vector's i i iii th component, we can also write this as
d x ∧ d y ( u , v ) = | d x ( u ) d x ( v ) d y ( u ) d y ( v ) | = | u 1 v 1 u 2 v 2 | = u 1 v 2 − u 2 v 1 d x ∧ d y ( u , v ) = d x ( u )      d x ( v ) d y ( u )      d y ( v ) = u 1 v 1 u 2 v 2 = u 1 v 2 − u 2 v 1 dx^^dy(u,v)=|[dx(u),dx(v)],[dy(u),dy(v)]|=|[u^(1),v^(1)],[u^(2),v^(2)]|=u^(1)v^(2)-u^(2)v^(1)d x \wedge d y(\mathbf{u}, \mathbf{v})=\left|\begin{array}{ll} d x(\mathbf{u}) & d x(\mathbf{v}) \\ d y(\mathbf{u}) & d y(\mathbf{v}) \end{array}\right|=\left|\begin{array}{cc} u^{1} & v^{1} \\ u^{2} & v^{2} \end{array}\right|=u^{1} v^{2}-u^{2} v^{1}dx∧dy(u,v)=|dx(u)dx(v)dy(u)dy(v)|=|u1v1u2v2|=u1v2−u2v1
where we define the signed area of the parallelogram spanned by u u u\mathbf{u}u and v v v\mathbf{v}v to be the value of d x ∧ d y ( u , v ) d x ∧ d y ( u , v ) dx^^dy(u,v)d x \wedge d y(\mathbf{u}, \mathbf{v})dx∧dy(u,v).
Earlier, in section 2.2 .3 , when looking at 1 -forms, we saw that d x i ( v ) d x i ( v ) dx^(i)(v)d x^{i}(\mathbf{v})dxi(v) gives the projection of vector v v v\mathbf{v}v onto the x i x i x^(i)x^{i}xi th coordinate axis. We can also make use of the geometric notion of projection in order to understand how higher degree forms act on vectors. We've just seen that for two vectors u u u\mathbf{u}u and v v v\mathbf{v}v in R 2 , d x ∧ d y ( u , v ) R 2 , d x ∧ d y ( u , v ) R^(2),dx^^dy(u,v)\mathbb{R}^{2}, d x \wedge d y(\mathbf{u}, \mathbf{v})R2,dx∧dy(u,v) gives the
signed area of the parallelogram spanned by u u u\mathbf{u}u and v v v\mathbf{v}v in R 2 R 2 R^(2)\mathbb{R}^{2}R2 itself. Therefore, d x ∧ d y d x ∧ d y dx^^dyd x \wedge d ydx∧dy projects onto the whole space of R 2 R 2 R^(2)\mathbb{R}^{2}R2.
Figure 2.7: d x ∧ d y d x ∧ d y dx^^dyd x \wedge d ydx∧dy acting on vectors u u u\mathbf{u}u and v v v\mathbf{v}v in R 3 R 3 R^(3)\mathbb{R}^{3}R3.
Now let's see how see how this works in R 3 R 3 R^(3)\mathbb{R}^{3}R3 with vectors u = ( u 1 , u 2 , u 3 ) u = u 1 , u 2 , u 3 u=(u^(1),u^(2),u^(3))\mathbf{u}=\left(u^{1}, u^{2}, u^{3}\right)u=(u1,u2,u3) and v = v = v=\mathbf{v}=v= ( v 1 , v 2 , v 3 ) v 1 , v 2 , v 3 (v^(1),v^(2),v^(3))\left(v^{1}, v^{2}, v^{3}\right)(v1,v2,v3), as shown in Figure 2.7 .
Again using the basis 1-forms d x d x dxd xdx and d y d y dyd ydy, we can define two new vectors on the x y x y xyx yxy plane: w 1 w 1 w_(1)\mathbf{w}_{1}w1 with components d x ( u ) = u 1 d x ( u ) = u 1 dx(u)=u^(1)d x(\mathbf{u})=u^{1}dx(u)=u1 and d y ( u ) = u 2 d y ( u ) = u 2 dy(u)=u^(2)d y(\mathbf{u})=u^{2}dy(u)=u2, and w 2 w 2 w_(2)\mathbf{w}_{2}w2 with components d x ( v ) = v 1 d x ( v ) = v 1 dx(v)=v^(1)d x(\mathbf{v})=v^{1}dx(v)=v1 and d y ( v ) = v 2 d y ( v ) = v 2 dy(v)=v^(2)d y(\mathbf{v})=v^{2}dy(v)=v2. The value of d x ∧ d y ( u , v ) d x ∧ d y ( u , v ) dx^^dy(u,v)d x \wedge d y(\mathbf{u}, \mathbf{v})dx∧dy(u,v) then equals the signed area of the parallelogram spanned by w 1 w 1 w_(1)\mathbf{w}_{1}w1 and w 2 w 2 w_(2)\mathbf{w}_{2}w2.
In other words, the parallelogram spanned by w 1 w 1 w_(1)\mathbf{w}_{1}w1 and w 2 w 2 w_(2)\mathbf{w}_{2}w2 is the projection onto the x y x y xyx yxy plane of the parallelogram spanned by u u u\mathbf{u}u and v v v\mathbf{v}v. The 2 -form d x ∧ d y d x ∧ d y dx^^dyd x \wedge d ydx∧dy allows us to find the area of that projected parallelogram, which is given by d x ∧ d y ( u , v ) d x ∧ d y ( u , v ) dx^^dy(u,v)d x \wedge d y(\mathbf{u}, \mathbf{v})dx∧dy(u,v). The area of the projected parallelogram in the x z x z xzx zxz plane would be given by d x ∧ d z ( u , v ) d x ∧ d z ( u , v ) dx^^dz(u,v)d x \wedge d z(\mathbf{u}, \mathbf{v})dx∧dz(u,v). And the area of the projected parallelogram in the y z y z yzy zyz plane would be given by d y ∧ d z ( u , v ) d y ∧ d z ( u , v ) dy^^dz(u,v)d y \wedge d z(\mathbf{u}, \mathbf{v})dy∧dz(u,v).
It's easy to see that if we used a d x ∧ d y a d x ∧ d y adx^^dya d x \wedge d yadx∧dy (where a a aaa is a constant) instead of d x ∧ d y d x ∧ d y dx^^dyd x \wedge d ydx∧dy, the value of a d x ∧ d z ( u , v ) a d x ∧ d z ( u , v ) adx^^dz(u,v)a d x \wedge d z(\mathbf{u}, \mathbf{v})adx∧dz(u,v) would be given by
(2.3.2) a d x ∧ d y ( u , v ) = a | d x ( u ) d x ( v ) d y ( u ) d y ( v ) | = a | u 1 v 1 u 2 v 2 | = a ( u 1 v 2 − u 2 v 1 ) (2.3.2) a d x ∧ d y ( u , v ) = a d x ( u ) d x ( v ) d y ( u ) d y ( v ) = a u 1 v 1 u 2 v 2 = a u 1 v 2 − u 2 v 1 {:(2.3.2)adx^^dy(u","v)=a|[dx(u),dx(v)],[dy(u),dy(v)]|=a|[u^(1),v^(1)],[u^(2),v^(2)]|=a(u^(1)v^(2)-u^(2)v^(1)):}a d x \wedge d y(\mathbf{u}, \mathbf{v})=a\left|\begin{array}{ll} d x(\mathbf{u}) & d x(\mathbf{v}) \tag{2.3.2}\\ d y(\mathbf{u}) & d y(\mathbf{v}) \end{array}\right|=a\left|\begin{array}{ll} u^{1} & v^{1} \\ u^{2} & v^{2} \end{array}\right|=a\left(u^{1} v^{2}-u^{2} v^{1}\right)(2.3.2)adx∧dy(u,v)=a|dx(u)dx(v)dy(u)dy(v)|=a|u1v1u2v2|=a(u1v2−u2v1)
which can be interpreted as the area of the projected parallelogram in the x y x y xyx yxy plane multiplied by the factor a a aaa.
Similarly, if we had a 2 -form ω = a d y ∧ d z + b d z ∧ d x + c d x ∧ d y ω = a d y ∧ d z + b d z ∧ d x + c d x ∧ d y omega=ady^^dz+bdz^^dx+cdx^^dy\omega=a d y \wedge d z+b d z \wedge d x+c d x \wedge d yω=ady∧dz+bdz∧dx+cdx∧dy, the value of ω ω omega\omegaω acting on u u u\mathbf{u}u and v v v\mathbf{v}v can be visualised as the area of the projected parallelogram in the y z y z yzy zyz plane multiplied by the factor a a aaa, plus the area of the projected parallelogram in the z x z x zxz xzx plane multiplied by the factor b b bbb, plus the area of the projected parallelogram in the x y x y xyx yxy plane multiplied by the factor c c ccc.
Example 2.3. From Bryan [7]. Find the value of the 2 -form ω = ( x + 2 z ) d x ∧ d y + y d x ∧ d z ω = ( x + 2 z ) d x ∧ d y + y d x ∧ d z omega=(x+2z)dx^^dy+ydx^^dz\boldsymbol{\omega}=(x+2 z) d x \wedge d y+y d x \wedge d zω=(x+2z)dx∧dy+ydx∧dz acting on the vectors v 1 = ( 1 , 3 , 3 ) v 1 = ( 1 , 3 , 3 ) v_(1)=(1,3,3)\mathbf{v}_{1}=(1,3,3)v1=(1,3,3) and v 2 = ( − 1 , 0 , 7 ) v 2 = ( − 1 , 0 , 7 ) v_(2)=(-1,0,7)\mathbf{v}_{2}=(-1,0,7)v2=(−1,0,7) at the point ( 1 , 1 , 5 ) ( 1 , 1 , 5 ) (1,1,5)(1,1,5)(1,1,5).
So we need to evaluate ω ( v 1 , v 2 ) ω v 1 , v 2 omega(v_(1),v_(2))\omega\left(\mathbf{v}_{1}, \mathbf{v}_{2}\right)ω(v1,v2) at the point ( 1 , 1 , 5 ) ( 1 , 1 , 5 ) (1,1,5)(1,1,5)(1,1,5). Utilising (2.3.2), we can write
ω ( v 1 , v 2 ) = ( ( x + 2 z ) d x ∧ d y + y d x ∧ d z ) ( v 1 , v 2 ) = ( x + 2 z ) | d x ( v 1 ) d x ( v 2 ) d y ( v 1 ) d y ( v 2 ) | + y | d x ( v 1 ) d x ( v 2 ) d z ( v 1 ) d z ( v 2 ) | = ( x + 2 z ) | d x ( 1 , 3 , 3 ) d x ( − 1 , 0 , 7 ) d y ( 1 , 3 , 3 ) d y ( − 1 , 0 , 7 ) | + y | d x ( 1 , 3 , 3 ) d x ( − 1 , 0 , 7 ) d z ( 1 , 3 , 3 ) d z ( − 1 , 0 , 7 ) | = ( x + 2 z ) | 1 − 1 3 0 | + y | 1 − 1 3 7 | = 3 ( x + 2 z ) + 10 y = 3 x + 10 y + 6 z ω v 1 , v 2 = ( ( x + 2 z ) d x ∧ d y + y d x ∧ d z ) v 1 , v 2 = ( x + 2 z ) d x v 1 d x v 2 d y v 1 d y v 2 + y d x v 1 d x v 2 d z v 1 d z v 2 = ( x + 2 z ) d x ( 1 , 3 , 3 ) d x ( − 1 , 0 , 7 ) d y ( 1 , 3 , 3 ) d y ( − 1 , 0 , 7 ) + y d x ( 1 , 3 , 3 ) d x ( − 1 , 0 , 7 ) d z ( 1 , 3 , 3 ) d z ( − 1 , 0 , 7 ) = ( x + 2 z ) 1 − 1 3 0 + y 1 − 1 3 7 = 3 ( x + 2 z ) + 10 y = 3 x + 10 y + 6 z {:[omega(v_(1),v_(2))=((x+2z)dx^^dy+ydx^^dz)(v_(1),v_(2))],[=(x+2z)|[dx(v_(1)),dx(v_(2))],[dy(v_(1)),dy(v_(2))]|+y|[dx(v_(1)),dx(v_(2))],[dz(v_(1)),dz(v_(2))]|],[=(x+2z)|[dx(1","3","3),dx(-1","0","7)],[dy(1","3","3),dy(-1","0","7)]|+y|[dx(1","3","3),dx(-1","0","7)],[dz(1","3","3),dz(-1","0","7)]|],[=(x+2z)|[1,-1],[3,0]|+y|[1,-1],[3,7]|],[=3(x+2z)+10 y],[=3x+10 y+6z]:}\begin{gathered} \omega\left(\mathbf{v}_{1}, \mathbf{v}_{2}\right)=((x+2 z) d x \wedge d y+y d x \wedge d z)\left(\mathbf{v}_{1}, \mathbf{v}_{2}\right) \\ =(x+2 z)\left|\begin{array}{ll} d x\left(\mathbf{v}_{1}\right) & d x\left(\mathbf{v}_{2}\right) \\ d y\left(\mathbf{v}_{1}\right) & d y\left(\mathbf{v}_{2}\right) \end{array}\right|+y\left|\begin{array}{ll} d x\left(\mathbf{v}_{1}\right) & d x\left(\mathbf{v}_{2}\right) \\ d z\left(\mathbf{v}_{1}\right) & d z\left(\mathbf{v}_{2}\right) \end{array}\right| \\ =(x+2 z)\left|\begin{array}{ll} d x(1,3,3) & d x(-1,0,7) \\ d y(1,3,3) & d y(-1,0,7) \end{array}\right|+y\left|\begin{array}{cc} d x(1,3,3) & d x(-1,0,7) \\ d z(1,3,3) & d z(-1,0,7) \end{array}\right| \\ =(x+2 z)\left|\begin{array}{cc} 1 & -1 \\ 3 & 0 \end{array}\right|+y\left|\begin{array}{cc} 1 & -1 \\ 3 & 7 \end{array}\right| \\ =3(x+2 z)+10 y \\ =3 x+10 y+6 z \end{gathered}ω(v1,v2)=((x+2z)dx∧dy+ydx∧dz)(v1,v2)=(x+2z)|dx(v1)dx(v2)dy(v1)dy(v2)|+y|dx(v1)dx(v2)dz(v1)dz(v2)|=(x+2z)|dx(1,3,3)dx(−1,0,7)dy(1,3,3)dy(−1,0,7)|+y|dx(1,3,3)dx(−1,0,7)dz(1,3,3)dz(−1,0,7)|=(x+2z)|1−130|+y|1−137|=3(x+2z)+10y=3x+10y+6z
At the point ( 1 , 1 , 5 ) ( 1 , 1 , 5 ) (1,1,5)(1,1,5)(1,1,5) this equals 43 .
Remember the determinant! We saw in section 1.7 that determinants are alternating multilinear functions of their columns (or rows). Because we naturally use determinants to calculate wedge products, those two properties also transfer over to differential forms. Feel free to convince yourself of this using the previous example:
  • Alternating - If we reverse the order of the two vectors and evaluate ω ( v 2 , v 1 ) ω v 2 , v 1 omega(v_(2),v_(1))\omega\left(\mathbf{v}_{2}, \mathbf{v}_{1}\right)ω(v2,v1) instead of ω ( v 1 , v 2 ) ω v 1 , v 2 omega(v_(1),v_(2))\omega\left(\mathbf{v}_{1}, \mathbf{v}_{2}\right)ω(v1,v2), we end up with -43 instead of 43 .
  • Multilinearity - If we multiply one of the vectors, v 2 v 2 v_(2)\mathbf{v}_{2}v2 for example, by a constant k k kkk and evaluate ω ( v 1 , k v 2 ) ω v 1 , k v 2 omega(v_(1),kv_(2))\omega\left(\mathbf{v}_{1}, k \mathbf{v}_{2}\right)ω(v1,kv2) instead of ω ( v 1 , v 2 ) ω v 1 , v 2 omega(v_(1),v_(2))\omega\left(\mathbf{v}_{1}, \mathbf{v}_{2}\right)ω(v1,v2), we end up with 43 k 43 k 43 k43 k43k instead of 43.
In general, the wedge product of two 1-forms ω 1 ω 1 omega_(1)\omega_{1}ω1 and ω 2 ω 2 omega_(2)\omega_{2}ω2 acting on two vectors v 1 v 1 v_(1)\mathbf{v}_{1}v1 and v 2 v 2 v_(2)\mathbf{v}_{2}v2 is given by
(2.3.3) ( ω 1 ∧ ω 2 ) ( v 1 , v 2 ) = | ω 1 ( v 1 ) ω 1 ( v 2 ) ω 2 ( v 1 ) ω 2 ( v 2 ) | (2.3.3) ω 1 ∧ ω 2 v 1 , v 2 = ω 1 v 1 ω 1 v 2 ω 2 v 1 ω 2 v 2 {:(2.3.3)(omega_(1)^^omega_(2))(v_(1),v_(2))=|[omega_(1)(v_(1)),omega_(1)(v_(2))],[omega_(2)(v_(1)),omega_(2)(v_(2))]|:}\left(\omega_{1} \wedge \omega_{2}\right)\left(\mathbf{v}_{1}, \mathbf{v}_{2}\right)=\left|\begin{array}{ll} \omega_{1}\left(\mathbf{v}_{1}\right) & \omega_{1}\left(\mathbf{v}_{2}\right) \tag{2.3.3}\\ \omega_{2}\left(\mathbf{v}_{1}\right) & \omega_{2}\left(\mathbf{v}_{2}\right) \end{array}\right|(2.3.3)(ω1∧ω2)(v1,v2)=|ω1(v1)ω1(v2)ω2(v1)ω2(v2)|

2.3.3 2-form example

Figure 2.8: Flux through a surface is a 2-form.
Working in R 3 R 3 R^(3)\mathbb{R}^{3}R3, Figure 2.8 shows a surface, the shaded parallelogram, spanned by u u u\mathbf{u}u and v v v\mathbf{v}v. If F F F\mathbf{F}F represents constant fluid flow through that surface, then the flux (flow per unit time) of the fluid through the shaded area is equal to the volume of the parallelepiped spanned by u , v u , v u,v\mathbf{u}, \mathbf{v}u,v and F F F\mathbf{F}F, ie by the determinant
| F , u , v | = | F 1 u 1 v 1 F 2 u 2 v 2 F 3 u 3 v 3 | | F , u , v | = F 1 u 1 v 1 F 2 u 2 v 2 F 3 u 3 v 3 |F,u,v|=|[F_(1),u^(1),v^(1)],[F_(2),u^(2),v^(2)],[F_(3),u^(3),v^(3)]||\mathbf{F}, \mathbf{u}, \mathbf{v}|=\left|\begin{array}{ccc} F_{1} & u^{1} & v^{1} \\ F_{2} & u^{2} & v^{2} \\ F_{3} & u^{3} & v^{3} \end{array}\right||F,u,v|=|F1u1v1F2u2v2F3u3v3|
which also happens to be the scalar triple product
F ⋅ ( u × v ) = | F , u , v | = | F 1 u 1 v 1 F 2 u 2 v 2 F 3 u 3 v 3 | F ⋅ ( u × v ) = | F , u , v | = F 1 u 1 v 1 F 2 u 2 v 2 F 3 u 3 v 3 F*(uxxv)=|F,u,v|=|[F_(1),u^(1),v^(1)],[F_(2),u^(2),v^(2)],[F_(3),u^(3),v^(3)]|\mathbf{F} \cdot(\mathbf{u} \times \mathbf{v})=|\mathbf{F}, \mathbf{u}, \mathbf{v}|=\left|\begin{array}{ccc} F_{1} & u^{1} & v^{1} \\ F_{2} & u^{2} & v^{2} \\ F_{3} & u^{3} & v^{3} \end{array}\right|F⋅(u×v)=|F,u,v|=|F1u1v1F2u2v2F3u3v3|
We can also describe the flux in terms of a 2 -form ω ω omega\omegaω acting on u u u\mathbf{u}u and v v v\mathbf{v}v, where
(2.3.4) ω ( u , v ) = F ⋅ ( u × v ) = | F , u , v | = | F 1 u 1 v 1 F 2 u 2 v 2 F 3 u 3 v 3 | (2.3.4) ω ( u , v ) = F ⋅ ( u × v ) = | F , u , v | = F 1 u 1 v 1 F 2 u 2 v 2 F 3 u 3 v 3 {:(2.3.4)omega(u","v)=F*(uxxv)=|F","u","v|=|[F_(1),u^(1),v^(1)],[F_(2),u^(2),v^(2)],[F_(3),u^(3),v^(3)]|:}\omega(\mathbf{u}, \mathbf{v})=\mathbf{F} \cdot(\mathbf{u} \times \mathbf{v})=|\mathbf{F}, \mathbf{u}, \mathbf{v}|=\left|\begin{array}{ccc} F_{1} & u^{1} & v^{1} \tag{2.3.4}\\ F_{2} & u^{2} & v^{2} \\ F_{3} & u^{3} & v^{3} \end{array}\right|(2.3.4)ω(u,v)=F⋅(u×v)=|F,u,v|=|F1u1v1F2u2v2F3u3v3|
The 2-form ω ω omega\omegaω is called the flux form of F F F\mathbf{F}F.
In R 3 R 3 R^(3)\mathbb{R}^{3}R3 all 2-forms are flux forms of some vector field.

2.3.4 Surfaces

A surface is a two-dimensional manifold and can be parameterised by two parameters, u u uuu and v v vvv for example. So in R 3 R 3 R^(3)\mathbb{R}^{3}R3 we might have a surface (in this case a cone open along
the x x xxx axis) described by the parametric equation
Φ ( u , v ) = ( u , u cos ⁡ v , u sin ⁡ v ) Φ ( u , v ) = ( u , u cos ⁡ v , u sin ⁡ v ) Phi(u,v)=(u,u cos v,u sin v)\boldsymbol{\Phi}(u, v)=(u, u \cos v, u \sin v)Φ(u,v)=(u,ucos⁡v,usin⁡v)
which means
x = u , y = u cos ⁡ v , z = u sin ⁡ v x = u , y = u cos ⁡ v , z = u sin ⁡ v x=u,y=u cos v,z=u sin vx=u, y=u \cos v, z=u \sin vx=u,y=ucos⁡v,z=usin⁡v
Tangent vectors to this surface are given by
∂ Φ ∂ u = ( ∂ x ∂ u , ∂ y ∂ u , ∂ z ∂ u ) = ( 1 , cos ⁡ v , sin ⁡ v ) ∂ Φ ∂ u = ∂ x ∂ u , ∂ y ∂ u , ∂ z ∂ u = ( 1 , cos ⁡ v , sin ⁡ v ) (del Phi)/(del u)=((del x)/(del u),(del y)/(del u),(del z)/(del u))=(1,cos v,sin v)\frac{\partial \Phi}{\partial u}=\left(\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u}\right)=(1, \cos v, \sin v)∂Φ∂u=(∂x∂u,∂y∂u,∂z∂u)=(1,cos⁡v,sin⁡v)
and
∂ Φ ∂ v = ( ∂ x ∂ v , ∂ y ∂ v , ∂ z ∂ v ) = ( 0 , − u sin ⁡ v , u cos ⁡ v ) ∂ Φ ∂ v = ∂ x ∂ v , ∂ y ∂ v , ∂ z ∂ v = ( 0 , − u sin ⁡ v , u cos ⁡ v ) (del Phi)/(del v)=((del x)/(del v),(del y)/(del v),(del z)/(del v))=(0,-u sin v,u cos v)\frac{\partial \boldsymbol{\Phi}}{\partial v}=\left(\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}, \frac{\partial z}{\partial v}\right)=(0,-u \sin v, u \cos v)∂Φ∂v=(∂x∂v,∂y∂v,∂z∂v)=(0,−usin⁡v,ucos⁡v)
We can show a 2-form ω ω omega\omegaω acting on the tangent vectors ∂ Φ ∂ u ∂ Φ ∂ u (del Phi)/(del u)\frac{\partial \Phi}{\partial u}∂Φ∂u and ∂ Φ ∂ v ∂ Φ ∂ v (del Phi)/(del v)\frac{\partial \Phi}{\partial v}∂Φ∂v by
ω ( ∂ Φ ∂ u , ∂ Φ ∂ v ) ω ∂ Φ ∂ u , ∂ Φ ∂ v omega((del Phi)/(del u),(del Phi)/(del v))\omega\left(\frac{\partial \Phi}{\partial u}, \frac{\partial \boldsymbol{\Phi}}{\partial v}\right)ω(∂Φ∂u,∂Φ∂v)
Earlier, when we were looking at how a 2 -form acts on two vectors, we noted that ( 2.3 .3 ) ( 2.3 .3 ) (2.3.3)(2.3 .3)(2.3.3)
( ω 1 ∧ ω 2 ) ( v 1 , v 2 ) = | ω 1 ( v 1 ) ω 1 ( v 2 ) ω 2 ( v 1 ) ω 2 ( v 2 ) | ω 1 ∧ ω 2 v 1 , v 2 = ω 1 v 1      ω 1 v 2 ω 2 v 1      ω 2 v 2 (omega_(1)^^omega_(2))(v_(1),v_(2))=|[omega_(1)(v_(1)),omega_(1)(v_(2))],[omega_(2)(v_(1)),omega_(2)(v_(2))]|\left(\omega_{1} \wedge \omega_{2}\right)\left(\mathbf{v}_{1}, \mathbf{v}_{2}\right)=\left|\begin{array}{ll} \omega_{1}\left(\mathbf{v}_{1}\right) & \omega_{1}\left(\mathbf{v}_{2}\right) \\ \omega_{2}\left(\mathbf{v}_{1}\right) & \omega_{2}\left(\mathbf{v}_{2}\right) \end{array}\right|(ω1∧ω2)(v1,v2)=|ω1(v1)ω1(v2)ω2(v1)ω2(v2)|
which works equally well if we write v 1 v 1 v_(1)\mathbf{v}_{1}v1 and v 2 v 2 v_(2)\mathbf{v}_{2}v2 explicitly as tangent vectors. For example, if we let ω 1 = d x ω 1 = d x omega_(1)=dx\omega_{1}=d xω1=dx and ω 2 = d y ω 2 = d y omega_(2)=dy\omega_{2}=d yω2=dy, and substitute ∂ Φ ∂ u ∂ Φ ∂ u (del Phi)/(del u)\frac{\partial \Phi}{\partial u}∂Φ∂u for v 1 v 1 v_(1)\mathbf{v}_{1}v1 and ∂ Φ ∂ v ∂ Φ ∂ v (del Phi)/(del v)\frac{\partial \Phi}{\partial v}∂Φ∂v for v 2 v 2 v_(2)\mathbf{v}_{2}v2, we obtain
( d x ∧ d y ) ( ∂ Φ ∂ u , ∂ Φ ∂ v ) = | d x ( ∂ Φ ∂ u ) d x ( ∂ Φ ∂ u ) d y ( ∂ Φ ∂ u ) d y ( ∂ Φ ∂ v ) | (2.3.5) ( d x ∧ d y ) ( ∂ Φ ∂ u , ∂ Φ ∂ v ) = | ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v | ( d x ∧ d y ) ∂ Φ ∂ u , ∂ Φ ∂ v = d x ∂ Φ ∂ u d x ∂ Φ ∂ u d y ∂ Φ ∂ u d y ∂ Φ ∂ v (2.3.5) ( d x ∧ d y ) ∂ Φ ∂ u , ∂ Φ ∂ v = ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v {:[(dx^^dy)((del Phi)/(del u),(del Phi)/(del v))=|[dx((del Phi)/(del u)),dx((del Phi)/(del u))],[dy((del Phi)/(del u)),dy((del Phi)/(del v))]|],[(2.3.5)(dx^^dy)((del Phi)/(del u),(del Phi)/(del v))=|[(del x)/(del u),(del x)/(del v)],[(del y)/(del u),(del y)/(del v)]|]:}\begin{gather*} (d x \wedge d y)\left(\frac{\partial \Phi}{\partial u}, \frac{\partial \Phi}{\partial v}\right)=\left|\begin{array}{ll} d x\left(\frac{\partial \Phi}{\partial u}\right) & d x\left(\frac{\partial \Phi}{\partial u}\right) \\ d y\left(\frac{\partial \Phi}{\partial u}\right) & d y\left(\frac{\partial \Phi}{\partial v}\right) \end{array}\right| \\ (d x \wedge d y)\left(\frac{\partial \Phi}{\partial u}, \frac{\partial \Phi}{\partial v}\right)=\left|\begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array}\right| \tag{2.3.5} \end{gather*}(dx∧dy)(∂Φ∂u,∂Φ∂v)=|dx(∂Φ∂u)dx(∂Φ∂u)dy(∂Φ∂u)dy(∂Φ∂v)|(2.3.5)(dx∧dy)(∂Φ∂u,∂Φ∂v)=|∂x∂u∂x∂v∂y∂u∂y∂v|
Similarly, we find
( d y ∧ d z ) ( ∂ Φ ∂ u , ∂ Φ ∂ v ) = | ∂ y ∂ u ∂ y ∂ v ∂ z ∂ u ∂ z ∂ v | ( d y ∧ d z ) ∂ Φ ∂ u , ∂ Φ ∂ v = ∂ y ∂ u      ∂ y ∂ v ∂ z ∂ u      ∂ z ∂ v (dy^^dz)((del Phi)/(del u),(del Phi)/(del v))=|[(del y)/(del u),(del y)/(del v)],[(del z)/(del u),(del z)/(del v)]|(d y \wedge d z)\left(\frac{\partial \Phi}{\partial u}, \frac{\partial \Phi}{\partial v}\right)=\left|\begin{array}{ll} \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} \end{array}\right|(dy∧dz)(∂Φ∂u,∂Φ∂v)=|∂y∂u∂y∂v∂z∂u∂z∂v|
and
( d z ∧ d x ) ( ∂ Φ ∂ u , ∂ Φ ∂ v ) = | ∂ z ∂ u ∂ z ∂ v ∂ x ∂ u ∂ x ∂ v | ( d z ∧ d x ) ∂ Φ ∂ u , ∂ Φ ∂ v = ∂ z ∂ u      ∂ z ∂ v ∂ x ∂ u      ∂ x ∂ v (dz^^dx)((del Phi)/(del u),(del Phi)/(del v))=|[(del z)/(del u),(del z)/(del v)],[(del x)/(del u),(del x)/(del v)]|(d z \wedge d x)\left(\frac{\partial \boldsymbol{\Phi}}{\partial u}, \frac{\partial \boldsymbol{\Phi}}{\partial v}\right)=\left|\begin{array}{ll} \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} \\ \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \end{array}\right|(dz∧dx)(∂Φ∂u,∂Φ∂v)=|∂z∂u∂z∂v∂x∂u∂x∂v|

2.4 3-forms and higher

We've just seen that a 2-form is a multilinear alternating function of two tangent vectors. Likewise, a 3 -form is an alternating multilinear function of three tangent vectors, and a k k kkk-form (where k > 3 k > 3 k > 3k>3k>3 ) is an alternating multilinear function of k k kkk tangent vectors.
On R 3 R 3 R^(3)\mathbb{R}^{3}R3 a 3-form will be a function of the basis 3 -form d x ∧ d y ∧ d z d x ∧ d y ∧ d z dx^^dy^^dzd x \wedge d y \wedge d zdx∧dy∧dz :
ω = f ( x , y , z ) d x ∧ d y ∧ d z ω = f ( x , y , z ) d x ∧ d y ∧ d z omega=f(x,y,z)dx^^dy^^dz\omega=f(x, y, z) d x \wedge d y \wedge d zω=f(x,y,z)dx∧dy∧dz
and can be constructed, using the wedge product, out of lower-order forms.
In this integral:
∫ M 5 z d x ∧ d y ∧ d z ⏟ 3 -form ∫ M   5 z d x ∧ d y ∧ d z ⏟ 3 -form  int_(M)ubrace(5zdx^^dy^^dzubrace)_(3"-form ")\int_{M} \underbrace{5 z d x \wedge d y \wedge d z}_{3 \text {-form }}∫M5zdx∧dy∧dz⏟3-form 
where M M MMM denotes a region of R 3 , 5 z d x ∧ d y ∧ d z R 3 , 5 z d x ∧ d y ∧ d z R^(3),5zdx^^dy^^dz\mathbb{R}^{3}, 5 z d x \wedge d y \wedge d zR3,5zdx∧dy∧dz is a 3 -form.
Again, because the wedge product is anti-commutative, the order of the differentials d x , d y d x , d y dx,dyd x, d ydx,dy and d z d z dzd zdz is not arbitrary.

2.4.1 3-forms acting on three vectors

Figure 2.9: d x ∧ d y ∧ d z d x ∧ d y ∧ d z dx^^dy^^dzd x \wedge d y \wedge d zdx∧dy∧dz acting on vectors u , v u , v u,v\mathbf{u}, \mathbf{v}u,v and w w w\mathbf{w}w in R 3 R 3 R^(3)\mathbb{R}^{3}R3.
A 3 -form acts on three vectors to give a number, ie a 3 -form is a function that requires three vector arguments. Let's see how this works in R 3 R 3 R^(3)\mathbb{R}^{3}R3 with the vectors u = ( u 1 , u 2 , u 3 ) u = u 1 , u 2 , u 3 u=(u^(1),u^(2),u^(3))\mathbf{u}=\left(u^{1}, u^{2}, u^{3}\right)u=(u1,u2,u3), v = ( v 1 , v 2 , v 3 ) v = v 1 , v 2 , v 3 v=(v^(1),v^(2),v^(3))\mathbf{v}=\left(v^{1}, v^{2}, v^{3}\right)v=(v1,v2,v3) and w = ( w 1 , w 2 , w 3 ) w = w 1 , w 2 , w 3 w=(w_(1),w_(2),w_(3))\mathbf{w}=\left(w_{1}, w_{2}, w_{3}\right)w=(w1,w2,w3), as shown in Figure 2.9. Earlier, when looking at the 2 -form d x ∧ d y d x ∧ d y dx^^dyd x \wedge d ydx∧dy, we said that a natural number defined from the two vectors u u u\mathbf{u}u and v v v\mathbf{v}v is the (signed) area of the parallelogram they span. which we then defined to be the value of d x ∧ d y ( u , v ) d x ∧ d y ( u , v ) dx^^dy(u,v)d x \wedge d y(\mathbf{u}, \mathbf{v})dx∧dy(u,v). In a similar fashion, we can define the value of the basis
3-form d x ∧ d y ∧ d z d x ∧ d y ∧ d z dx^^dy^^dzd x \wedge d y \wedge d zdx∧dy∧dz acting on the three vectors u , v u , v u,v\mathbf{u}, \mathbf{v}u,v and w w w\mathbf{w}w as the signed volume of the parallelepiped they span. This signed volume is given by the determinant
| u 1 v 1 w 1 u 2 v 2 w 2 u 3 v 3 w 3 | u 1 v 1 w 1 u 2 v 2 w 2 u 3 v 3 w 3 |[u^(1),v^(1),w^(1)],[u^(2),v^(2),w^(2)],[u^(3),v^(3),w^(3)]|\left|\begin{array}{ccc} u^{1} & v^{1} & w^{1} \\ u^{2} & v^{2} & w^{2} \\ u^{3} & v^{3} & w^{3} \end{array}\right||u1v1w1u2v2w2u3v3w3|
Again we recall that d x i d x i dx^(i)d x^{i}dxi picks out a vector's i i iii th component and justify our definition by writing
d x ∧ d y ∧ d z ( u , v , w ) = | d x ( u ) d x ( v ) d x ( w ) d y ( u ) d y ( v ) d y ( w ) d z ( u ) d z ( v ) d z ( w ) | = | u 1 v 1 w 1 u 2 v 2 w 2 u 3 v 3 w 3 | d x ∧ d y ∧ d z ( u , v , w ) = d x ( u )      d x ( v )      d x ( w ) d y ( u )      d y ( v )      d y ( w ) d z ( u )      d z ( v )      d z ( w ) = u 1 v 1 w 1 u 2 v 2 w 2 u 3 v 3 w 3 dx^^dy^^dz(u,v,w)=|[dx(u),dx(v),dx(w)],[dy(u),dy(v),dy(w)],[dz(u),dz(v),dz(w)]|=|[u^(1),v^(1),w^(1)],[u^(2),v^(2),w^(2)],[u^(3),v^(3),w^(3)]|d x \wedge d y \wedge d z(\mathbf{u}, \mathbf{v}, \mathbf{w})=\left|\begin{array}{lll} d x(\mathbf{u}) & d x(\mathbf{v}) & d x(\mathbf{w}) \\ d y(\mathbf{u}) & d y(\mathbf{v}) & d y(\mathbf{w}) \\ d z(\mathbf{u}) & d z(\mathbf{v}) & d z(\mathbf{w}) \end{array}\right|=\left|\begin{array}{ccc} u^{1} & v^{1} & w^{1} \\ u^{2} & v^{2} & w^{2} \\ u^{3} & v^{3} & w^{3} \end{array}\right|dx∧dy∧dz(u,v,w)=|dx(u)dx(v)dx(w)dy(u)dy(v)dy(w)dz(u)dz(v)dz(w)|=|u1v1w1u2v2w2u3v3w3|
Just as we did with the 2 -form d x ∧ d y d x ∧ d y dx^^dyd x \wedge d ydx∧dy, we can now ask what space does the 3 -form d x ∧ d y ∧ d z d x ∧ d y ∧ d z dx^^dy^^dzd x \wedge d y \wedge d zdx∧dy∧dz project u , v u , v u,v\mathbf{u}, \mathbf{v}u,v and w w w\mathbf{w}w onto in R 3 R 3 R^(3)\mathbb{R}^{3}R3 ? We've just seen that d x ∧ d y ∧ d z ( u , v , w ) d x ∧ d y ∧ d z ( u , v , w ) dx^^dy^^dz(u,v,w)d x \wedge d y \wedge d z(\mathbf{u}, \mathbf{v}, \mathbf{w})dx∧dy∧dz(u,v,w) gives the signed volume of the parallelepiped spanned by u , v u , v u,v\mathbf{u}, \mathbf{v}u,v and w w w\mathbf{w}w in R 3 R 3 R^(3)\mathbb{R}^{3}R3. Therefore, d x ∧ d y ∧ d z d x ∧ d y ∧ d z dx^^dy^^dzd x \wedge d y \wedge d zdx∧dy∧dz projects onto the whole space of R 3 R 3 R^(3)\mathbb{R}^{3}R3.
Now let's exercise our mathematical imaginations and explore what happens if we have a 3 -form d x 1 ∧ d x 2 ∧ d x 3 d x 1 ∧ d x 2 ∧ d x 3 dx^(1)^^dx^(2)^^dx^(3)d x^{1} \wedge d x^{2} \wedge d x^{3}dx1∧dx2∧dx3 acting on three vectors v 1 = ( 1 , 2 , − 1 , 1 ) , v 2 = ( 3 , − 2 , 1 , 2 ) v 1 = ( 1 , 2 , − 1 , 1 ) , v 2 = ( 3 , − 2 , 1 , 2 ) v_(1)=(1,2,-1,1),v_(2)=(3,-2,1,2)\mathbf{v}_{1}=(1,2,-1,1), \mathbf{v}_{2}=(3,-2,1,2)v1=(1,2,−1,1),v2=(3,−2,1,2) and v 3 = ( 0 , 1 , 2 , 1 ) v 3 = ( 0 , 1 , 2 , 1 ) v_(3)=(0,1,2,1)\mathbf{v}_{3}=(0,1,2,1)v3=(0,1,2,1) in R 4 R 4 R^(4)\mathbb{R}^{4}R4, using x 1 , x 2 , x 3 , x 4 x 1 , x 2 , x 3 , x 4 x^(1),x^(2),x^(3),x^(4)x^{1}, x^{2}, x^{3}, x^{4}x1,x2,x3,x4 coordinates. The vectors v 1 , v 2 v 1 , v 2 v_(1),v_(2)\mathbf{v}_{1}, \mathbf{v}_{2}v1,v2 and v 3 v 3 v_(3)\mathbf{v}_{3}v3 span a parallelepiped in R 4 R 4 R^(4)\mathbb{R}^{4}R4. Using d x 1 ∧ d x 2 ∧ d x 3 d x 1 ∧ d x 2 ∧ d x 3 dx^(1)^^dx^(2)^^dx^(3)d x^{1} \wedge d x^{2} \wedge d x^{3}dx1∧dx2∧dx3 we define three new vectors w 1 , w 2 w 1 , w 2 w_(1),w_(2)\mathbf{w}_{1}, \mathbf{w}_{2}w1,w2 and w 3 w 3 w_(3)\mathbf{w}_{3}w3 :
w 1 = d x 1 ∧ d x 2 ∧ d x 3 ( v 1 ) = ( 1 , 2 , − 1 ) w 2 = d x 1 ∧ d x 2 ∧ d x 3 ( v 2 ) = ( 3 , − 2 , 1 ) w 3 = d x 1 ∧ d x 2 ∧ d x 3 ( v 3 ) = ( 0 , 1 , 2 ) w 1 = d x 1 ∧ d x 2 ∧ d x 3 v 1 = ( 1 , 2 , − 1 ) w 2 = d x 1 ∧ d x 2 ∧ d x 3 v 2 = ( 3 , − 2 , 1 ) w 3 = d x 1 ∧ d x 2 ∧ d x 3 v 3 = ( 0 , 1 , 2 ) {:[w_(1)=dx^(1)^^dx^(2)^^dx^(3)(v_(1))=(1","2","-1)],[w_(2)=dx^(1)^^dx^(2)^^dx^(3)(v_(2))=(3","-2","1)],[w_(3)=dx^(1)^^dx^(2)^^dx^(3)(v_(3))=(0","1","2)]:}\begin{gathered} \mathbf{w}_{1}=d x^{1} \wedge d x^{2} \wedge d x^{3}\left(\mathbf{v}_{1}\right)=(1,2,-1) \\ \mathbf{w}_{2}=d x^{1} \wedge d x^{2} \wedge d x^{3}\left(\mathbf{v}_{2}\right)=(3,-2,1) \\ \mathbf{w}_{3}=d x^{1} \wedge d x^{2} \wedge d x^{3}\left(\mathbf{v}_{3}\right)=(0,1,2) \end{gathered}w1=dx1∧dx2∧dx3(v1)=(1,2,−1)w2=dx1∧dx2∧dx3(v2)=(3,−2,1)w3=dx1∧dx2∧dx3(v3)=(0,1,2)
The parallelepiped spanned by w 1 , w 2 w 1 , w 2 w_(1),w_(2)\mathbf{w}_{1}, \mathbf{w}_{2}w1,w2 and w 3 w 3 w_(3)\mathbf{w}_{3}w3 is the projection onto the x 1 x 2 x 3 x 1 x 2 x 3 x_(1)x_(2)x_(3)x_{1} x_{2} x_{3}x1x2x3 subspace of R 4 R 4 R^(4)\mathbb{R}^{4}R4 of the parallelepiped spanned by v 1 , v 2 v 1 , v 2 v_(1),v_(2)\mathbf{v}_{1}, \mathbf{v}_{2}v1,v2 and v 3 v 3 v_(3)\mathbf{v}_{3}v3. The value of x 1 ∧ d x 2 ∧ d x 3 ( v 1 , v 2 , v 3 ) x 1 ∧ d x 2 ∧ d x 3 v 1 , v 2 , v 3 x^(1)^^dx^(2)^^dx^(3)(v_(1),v_(2),v_(3))x^{1} \wedge d x^{2} \wedge d x^{3}\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right)x1∧dx2∧dx3(v1,v2,v3) then gives the volume of that projected parallelepiped, ie
d x 1 ∧ d x 2 ∧ d x 3 ( v 1 , v 2 , v 3 ) = | d x 1 ( v 1 ) d x 1 ( v 2 ) d x 1 ( v 3 ) d x 2 ( v 1 ) d x 2 ( v 2 ) d x 2 ( v 3 ) d x 3 ( v 1 ) d x 3 ( v 2 ) d x 3 ( v 3 ) | d x 1 ∧ d x 2 ∧ d x 3 v 1 , v 2 , v 3 = d x 1 v 1      d x 1 v 2      d x 1 v 3 d x 2 v 1      d x 2 v 2      d x 2 v 3 d x 3 v 1      d x 3 v 2      d x 3 v 3 dx^(1)^^dx^(2)^^dx^(3)(v_(1),v_(2),v_(3))=|[dx^(1)(v_(1)),dx^(1)(v_(2)),dx^(1)(v_(3))],[dx^(2)(v_(1)),dx^(2)(v_(2)),dx^(2)(v_(3))],[dx^(3)(v_(1)),dx^(3)(v_(2)),dx^(3)(v_(3))]|d x^{1} \wedge d x^{2} \wedge d x^{3}\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right)=\left|\begin{array}{lll} d x^{1}\left(\mathbf{v}_{1}\right) & d x^{1}\left(\mathbf{v}_{2}\right) & d x^{1}\left(\mathbf{v}_{3}\right) \\ d x^{2}\left(\mathbf{v}_{1}\right) & d x^{2}\left(\mathbf{v}_{2}\right) & d x^{2}\left(\mathbf{v}_{3}\right) \\ d x^{3}\left(\mathbf{v}_{1}\right) & d x^{3}\left(\mathbf{v}_{2}\right) & d x^{3}\left(\mathbf{v}_{3}\right) \end{array}\right|dx1∧dx2∧dx3(v1,v2,v3)=|dx1(v1)dx1(v2)dx1(v3)dx2(v1)dx2(v2)dx2(v3)dx3(v1)dx3(v2)dx3(v3)|
If we used a d x 1 ∧ d x 2 ∧ d x 3 a d x 1 ∧ d x 2 ∧ d x 3 adx^(1)^^dx^(2)^^dx^(3)a d x^{1} \wedge d x^{2} \wedge d x^{3}adx1∧dx2∧dx3 (where a a aaa is a constant) instead of d x 1 ∧ d x 2 ∧ d x 3 d x 1 ∧ d x 2 ∧ d x 3 dx^(1)^^dx^(2)^^dx^(3)d x^{1} \wedge d x^{2} \wedge d x^{3}dx1∧dx2∧dx3, the value of a d x 1 ∧ d x 2 ∧ d x 3 ( v 1 , v 2 , v 3 ) a d x 1 ∧ d x 2 ∧ d x 3 v 1 , v 2 , v 3 adx^(1)^^dx^(2)^^dx^(3)(v_(1),v_(2),v_(3))a d x^{1} \wedge d x^{2} \wedge d x^{3}\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right)adx1∧dx2∧dx3(v1,v2,v3) can be interpreted as the volume of the projected parallelepiped in the x 1 x 2 x 3 x 1 x 2 x 3 x_(1)x_(2)x_(3)x_{1} x_{2} x_{3}x1x2x3 subspace of R 4 R 4 R^(4)\mathbb{R}^{4}R4 multiplied by the factor a a aaa.
Similarly, if we had a general 3 -form in R 4 R 4 R^(4)\mathbb{R}^{4}R4
ω = a d x 1 ∧ d x 2 ∧ d x 3 + b d x 1 ∧ d x 2 ∧ d x 4 + c d x 1 ∧ d x 3 ∧ d x 4 + d d x 2 ∧ d x 3 ∧ d x 4 ω = a d x 1 ∧ d x 2 ∧ d x 3 + b d x 1 ∧ d x 2 ∧ d x 4 + c d x 1 ∧ d x 3 ∧ d x 4 + d d x 2 ∧ d x 3 ∧ d x 4 omega=adx^(1)^^dx^(2)^^dx^(3)+bdx^(1)^^dx^(2)^^dx^(4)+cdx^(1)^^dx^(3)^^dx^(4)+ddx^(2)^^dx^(3)^^dx^(4)\omega=a d x^{1} \wedge d x^{2} \wedge d x^{3}+b d x^{1} \wedge d x^{2} \wedge d x^{4}+c d x^{1} \wedge d x^{3} \wedge d x^{4}+d d x^{2} \wedge d x^{3} \wedge d x^{4}ω=adx1∧dx2∧dx3+bdx1∧dx2∧dx4+cdx1∧dx3∧dx4+ddx2∧dx3∧dx4
(where a , b , c a , b , c a,b,ca, b, ca,b,c and d d ddd are constants), the value of ω ω omega\omegaω acting on v 1 , v 2 v 1 , v 2 v_(1),v_(2)\mathbf{v}_{1}, \mathbf{v}_{2}v1,v2 and v 3 v 3 v_(3)\mathbf{v}_{3}v3 can be visualised as the area of the projected parallelogram in the x 1 x 2 x 3 x 1 x 2 x 3 x_(1)x_(2)x_(3)x_{1} x_{2} x_{3}x1x2x3 subspace multiplied by the factor a a aaa, plus the area of the projected parallelogram in the x 1 x 2 x 4 x 1 x 2 x 4 x_(1)x_(2)x_(4)x_{1} x_{2} x_{4}x1x2x4 subspace
multiplied by the factor b b bbb, plus the area of the projected parallelogram in the x 1 x 3 x 4 x 1 x 3 x 4 x_(1)x_(3)x_(4)x_{1} x_{3} x_{4}x1x3x4 subspace multiplied by the factor c c ccc, plus the area of the projected parallelogram in the x 2 x 3 x 4 x 2 x 3 x 4 x_(2)x_(3)x_(4)x_{2} x_{3} x_{4}x2x3x4 subspace multiplied by the factor d d ddd.
In general, the wedge product of three 1 -forms ω 1 , ω 2 ω 1 , ω 2 omega_(1),omega_(2)\omega_{1}, \omega_{2}ω1,ω2 and ω 3 ω 3 omega_(3)\omega_{3}ω3 acting on three vectors v 1 , v 2 v 1 , v 2 v_(1),v_(2)\mathbf{v}_{1}, \mathbf{v}_{2}v1,v2 and v 3 v 3 v_(3)\mathbf{v}_{3}v3 is given by
(2.4.1) ω 1 ∧ ω 2 ∧ ω 3 ( v 1 , v 2 , v 3 ) = | ω 1 ( v 1 ) ω 1 ( v 2 ) ω 1 ( v 3 ) ω 2 ( v 1 ) ω 2 ( v 2 ) ω 2 ( v 3 ) ω 3 ( v 1 ) ω 3 ( v 2 ) ω 3 ( v 3 ) | (2.4.1) ω 1 ∧ ω 2 ∧ ω 3 v 1 , v 2 , v 3 = ω 1 v 1 ω 1 v 2 ω 1 v 3 ω 2 v 1 ω 2 v 2 ω 2 v 3 ω 3 v 1 ω 3 v 2 ω 3 v 3 {:(2.4.1)omega_(1)^^omega_(2)^^omega_(3)(v_(1),v_(2),v_(3))=|[omega_(1)(v_(1)),omega_(1)(v_(2)),omega_(1)(v_(3))],[omega_(2)(v_(1)),omega_(2)(v_(2)),omega_(2)(v_(3))],[omega_(3)(v_(1)),omega_(3)(v_(2)),omega_(3)(v_(3))]|:}\omega_{1} \wedge \omega_{2} \wedge \omega_{3}\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right)=\left|\begin{array}{lll} \omega_{1}\left(\mathbf{v}_{1}\right) & \omega_{1}\left(\mathbf{v}_{2}\right) & \omega_{1}\left(\mathbf{v}_{3}\right) \tag{2.4.1}\\ \omega_{2}\left(\mathbf{v}_{1}\right) & \omega_{2}\left(\mathbf{v}_{2}\right) & \omega_{2}\left(\mathbf{v}_{3}\right) \\ \omega_{3}\left(\mathbf{v}_{1}\right) & \omega_{3}\left(\mathbf{v}_{2}\right) & \omega_{3}\left(\mathbf{v}_{3}\right) \end{array}\right|(2.4.1)ω1∧ω2∧ω3(v1,v2,v3)=|ω1(v1)ω1(v2)ω1(v3)ω2(v1)ω2(v2)ω2(v3)ω3(v1)ω3(v2)ω3(v3)|
To generalise (2.4.1), the wedge product of k 1 k 1 k1k 1k1-forms ω 1 , ⋯ , ω k ω 1 , ⋯ , ω k omega_(1),cdots,omega_(k)\omega_{1}, \cdots, \omega_{k}ω1,⋯,ωk acting on k k kkk vectors v 1 , … , v k v 1 , … , v k v_(1),dots,v_(k)\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}v1,…,vk is given by
( ω 1 ∧ ⋯ ∧ ω k ) ( v 1 , … , v k ) = | ω 1 ( v 1 ) ⋯ ω 1 ( v k ) ⋮ ⋮ ω k ( v 1 ) ⋯ ω k ( v k ) | ω 1 ∧ ⋯ ∧ ω k v 1 , … , v k = ω 1 v 1 ⋯ ω 1 v k ⋮ ⋮ ω k v 1 ⋯ ω k v k (omega_(1)^^cdots^^omega_(k))(v_(1),dots,v_(k))=|[omega_(1)(v_(1)),cdots,omega_(1)(v_(k))],[vdots,,vdots],[omega_(k)(v_(1)),cdots,omega_(k)(v_(k))]|\left(\omega_{1} \wedge \cdots \wedge \omega_{k}\right)\left(\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right)=\left|\begin{array}{ccc} \omega_{1}\left(\mathbf{v}_{1}\right) & \cdots & \omega_{1}\left(\mathbf{v}_{k}\right) \\ \vdots & & \vdots \\ \omega_{k}\left(\mathbf{v}_{1}\right) & \cdots & \omega_{k}\left(\mathbf{v}_{k}\right) \end{array}\right|(ω1∧⋯∧ωk)(v1,…,vk)=|ω1(v1)⋯ω1(vk)⋮⋮ωk(v1)⋯ωk(vk)|
Example 2.4. From Hubbard and Hubbard [11]. Find the value of the basis 3-form d x 1 ∧ d x 2 ∧ d x 4 d x 1 ∧ d x 2 ∧ d x 4 dx^(1)^^dx^(2)^^dx^(4)d x^{1} \wedge d x^{2} \wedge d x^{4}dx1∧dx2∧dx4 acting on the vectors v 1 = ( 1 , 2 , − 1 , 1 ) , v 2 = ( 3 , − 2 , 1 , 2 ) v 1 = ( 1 , 2 , − 1 , 1 ) , v 2 = ( 3 , − 2 , 1 , 2 ) v_(1)=(1,2,-1,1),v_(2)=(3,-2,1,2)\mathbf{v}_{1}=(1,2,-1,1), \mathbf{v}_{2}=(3,-2,1,2)v1=(1,2,−1,1),v2=(3,−2,1,2) and v 3 = ( 0 , 1 , 2 , 1 ) v 3 = ( 0 , 1 , 2 , 1 ) v_(3)=(0,1,2,1)\mathbf{v}_{3}=(0,1,2,1)v3=(0,1,2,1)
Equation ( 2.4 .1 ) ( 2.4 .1 ) (2.4.1)(2.4 .1)(2.4.1)
( ω 1 ∧ ω 2 ∧ ω 3 ) ( v 1 , v 2 , v 3 ) = | ω 1 ( v 1 ) ω 1 ( v 2 ) ω 1 ( v 3 ) ω 2 ( v 1 ) ω 2 ( v 2 ) ω 2 ( v 3 ) ω 3 ( v 1 ) ω 3 ( v 2 ) ω 3 ( v 3 ) | ω 1 ∧ ω 2 ∧ ω 3 v 1 , v 2 , v 3 = ω 1 v 1      ω 1 v 2      ω 1 v 3 ω 2 v 1      ω 2 v 2      ω 2 v 3 ω 3 v 1      ω 3 v 2      ω 3 v 3 (omega_(1)^^omega_(2)^^omega_(3))(v_(1),v_(2),v_(3))=|[omega_(1)(v_(1)),omega_(1)(v_(2)),omega_(1)(v_(3))],[omega_(2)(v_(1)),omega_(2)(v_(2)),omega_(2)(v_(3))],[omega_(3)(v_(1)),omega_(3)(v_(2)),omega_(3)(v_(3))]|\left(\omega_{1} \wedge \omega_{2} \wedge \omega_{3}\right)\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right)=\left|\begin{array}{lll} \omega_{1}\left(\mathbf{v}_{1}\right) & \omega_{1}\left(\mathbf{v}_{2}\right) & \omega_{1}\left(\mathbf{v}_{3}\right) \\ \omega_{2}\left(\mathbf{v}_{1}\right) & \omega_{2}\left(\mathbf{v}_{2}\right) & \omega_{2}\left(\mathbf{v}_{3}\right) \\ \omega_{3}\left(\mathbf{v}_{1}\right) & \omega_{3}\left(\mathbf{v}_{2}\right) & \omega_{3}\left(\mathbf{v}_{3}\right) \end{array}\right|(ω1∧ω2∧ω3)(v1,v2,v3)=|ω1(v1)ω1(v2)ω1(v3)ω2(v1)ω2(v2)ω2(v3)ω3(v1)ω3(v2)ω3(v3)|
becomes, for our example,
( d x 1 ∧ d x 2 ∧ d x 4 ) ( v 1 , v 2 , v 3 ) = | d x 1 ( v 1 ) d x 1 ( v 2 ) d x 1 ( v 3 ) d x 2 ( v 1 ) d x 2 ( v 2 ) d x 2 ( v 3 ) d x 4 ( v 1 ) d x 4 ( v 2 ) d x 4 ( v 3 ) | = | 1 3 0 2 − 2 1 1 2 1 | = − 7 . d x 1 ∧ d x 2 ∧ d x 4 v 1 , v 2 , v 3 = d x 1 v 1 d x 1 v 2 d x 1 v 3 d x 2 v 1 d x 2 v 2 d x 2 v 3 d x 4 v 1 d x 4 v 2 d x 4 v 3 = 1 3 0 2 − 2 1 1 2 1 = − 7 . {:[(dx^(1)^^dx^(2)^^dx^(4))(v_(1),v_(2),v_(3))=|[dx^(1)(v_(1)),dx^(1)(v_(2)),dx^(1)(v_(3))],[dx^(2)(v_(1)),dx^(2)(v_(2)),dx^(2)(v_(3))],[dx^(4)(v_(1)),dx^(4)(v_(2)),dx^(4)(v_(3))]|],[=|[1,3,0],[2,-2,1],[1,2,1]|],[=-7.]:}\begin{gathered} \left(d x^{1} \wedge d x^{2} \wedge d x^{4}\right)\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right)=\left|\begin{array}{lll} d x^{1}\left(\mathbf{v}_{1}\right) & d x^{1}\left(\mathbf{v}_{2}\right) & d x^{1}\left(\mathbf{v}_{3}\right) \\ d x^{2}\left(\mathbf{v}_{1}\right) & d x^{2}\left(\mathbf{v}_{2}\right) & d x^{2}\left(\mathbf{v}_{3}\right) \\ d x^{4}\left(\mathbf{v}_{1}\right) & d x^{4}\left(\mathbf{v}_{2}\right) & d x^{4}\left(\mathbf{v}_{3}\right) \end{array}\right| \\ =\left|\begin{array}{ccc} 1 & 3 & 0 \\ 2 & -2 & 1 \\ 1 & 2 & 1 \end{array}\right| \\ =-7 . \end{gathered}(dx1∧dx2∧dx4)(v1,v2,v3)=|dx1(v1)dx1(v2)dx1(v3)dx2(v1)dx2(v2)dx2(v3)dx4(v1)dx4(v2)dx4(v3)|=|1302−21121|=−7.
You might like to confirm the alternating and multilinear properties of the 3 -form d x 1 ∧ d x 2 ∧ d x 4 d x 1 ∧ d x 2 ∧ d x 4 dx^(1)^^dx^(2)^^dx^(4)d x^{1} \wedge d x^{2} \wedge d x^{4}dx1∧dx2∧dx4 in the previous example:
  • Alternating - If we exchange any two of the vectors in ( d x 1 ∧ d x 2 ∧ d x 4 ) ( v 1 , v 2 , v 3 ) d x 1 ∧ d x 2 ∧ d x 4 v 1 , v 2 , v 3 (dx^(1)^^dx^(2)^^dx^(4))(v_(1),v_(2),v_(3))\left(d x^{1} \wedge d x^{2} \wedge d x^{4}\right)\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right)(dx1∧dx2∧dx4)(v1,v2,v3), we end up with 7 instead of -7 .
  • Multilinear - If we multiply one of the vectors, v 2 v 2 v_(2)\mathbf{v}_{2}v2 for example, by a constant k k kkk and evaluate ( d x 1 ∧ d x 2 ∧ d x 4 ) ( v 1 , k v 2 , v 3 ) d x 1 ∧ d x 2 ∧ d x 4 v 1 , k v 2 , v 3 (dx^(1)^^dx^(2)^^dx^(4))(v_(1),kv_(2),v_(3))\left(d x^{1} \wedge d x^{2} \wedge d x^{4}\right)\left(\mathbf{v}_{1}, k \mathbf{v}_{2}, \mathbf{v}_{3}\right)(dx1∧dx2∧dx4)(v1,kv2,v3) instead of ( d x 1 ∧ d x 2 ∧ d x 4 ) ( v 1 , v 2 , v 3 ) d x 1 ∧ d x 2 ∧ d x 4 v 1 , v 2 , v 3 (dx^(1)^^dx^(2)^^dx^(4))(v_(1),v_(2),v_(3))\left(d x^{1} \wedge d x^{2} \wedge d x^{4}\right)\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right)(dx1∧dx2∧dx4)(v1,v2,v3), we end up with − 7 k − 7 k -7k-7 k−7k instead of -7 .

2.4.2 3-form example

Referring again to Figure 2.8, the oriented volume of the parallelepiped spanned by u , v u , v u,v\mathbf{u}, \mathbf{v}u,v and F F F\mathbf{F}F is a 3 -form ω ω omega\omegaω, where
ω ( F , u , v ) = | F 1 u 1 v 1 F 2 u 2 v 2 F 3 u 3 v 3 | ω ( F , u , v ) = F 1 u 1 v 1 F 2 u 2 v 2 F 3 u 3 v 3 omega(F,u,v)=|[F_(1),u^(1),v^(1)],[F_(2),u^(2),v^(2)],[F_(3),u^(3),v^(3)]|\omega(\mathbf{F}, \mathbf{u}, \mathbf{v})=\left|\begin{array}{ccc} F_{1} & u^{1} & v^{1} \\ F_{2} & u^{2} & v^{2} \\ F_{3} & u^{3} & v^{3} \end{array}\right|ω(F,u,v)=|F1u1v1F2u2v2F3u3v3|
In R 3 3 R 3 3 R^(3)3\mathbb{R}^{3} 3R33-forms are sometimes called density forms. If the function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x, y, z)f(x,y,z) in the general 3 -form
ω = f ( x , y , z ) d x ∧ d y ∧ d z ω = f ( x , y , z ) d x ∧ d y ∧ d z omega=f(x,y,z)dx^^dy^^dz\omega=f(x, y, z) d x \wedge d y \wedge d zω=f(x,y,z)dx∧dy∧dz
describes something (mass, for example) per unit volume - ie is a density function we can integrate ω ω omega\omegaω to find the total amount of that something.

2.4.3 Three and higher dimensional spaces

A three-dimensional manifold can be parameterised by three parameters, u , v u , v u,vu, vu,v and w w www for example, and described by a parametric equation Φ ( u , v , w ) Φ ( u , v , w ) Phi(u,v,w)\boldsymbol{\Phi}(u, v, w)Φ(u,v,w). If we have a three-dimensional manifold in R 4 R 4 R^(4)\mathbb{R}^{4}R4 (with coordinates x 1 , x 2 , x 3 , x 4 x 1 , x 2 , x 3 , x 4 x^(1),x^(2),x^(3),x^(4)x^{1}, x^{2}, x^{3}, x^{4}x1,x2,x3,x4 ), the tangent vectors are
∂ Φ ∂ u = ( ∂ x 1 ∂ u , ∂ x 2 ∂ u , ∂ x 3 ∂ u , ∂ x 4 ∂ u ) ∂ Φ ∂ v = ( ∂ x 1 ∂ v , ∂ x 2 ∂ v , ∂ x 3 ∂ v , ∂ x 4 ∂ v ) ∂ Φ ∂ u = ∂ x 1 ∂ u , ∂ x 2 ∂ u , ∂ x 3 ∂ u , ∂ x 4 ∂ u ∂ Φ ∂ v = ∂ x 1 ∂ v , ∂ x 2 ∂ v , ∂ x 3 ∂ v , ∂ x 4 ∂ v {:[(del Phi)/(del u)=((delx^(1))/(del u),(delx^(2))/(del u),(delx^(3))/(del u),(delx^(4))/(del u))],[(del Phi)/(del v)=((delx^(1))/(del v),(delx^(2))/(del v),(delx^(3))/(del v),(delx^(4))/(del v))]:}\begin{aligned} & \frac{\partial \boldsymbol{\Phi}}{\partial u}=\left(\frac{\partial x^{1}}{\partial u}, \frac{\partial x^{2}}{\partial u}, \frac{\partial x^{3}}{\partial u}, \frac{\partial x^{4}}{\partial u}\right) \\ & \frac{\partial \boldsymbol{\Phi}}{\partial v}=\left(\frac{\partial x^{1}}{\partial v}, \frac{\partial x^{2}}{\partial v}, \frac{\partial x^{3}}{\partial v}, \frac{\partial x^{4}}{\partial v}\right) \end{aligned}∂Φ∂u=(∂x1∂u,∂x2∂u,∂x3∂u,∂x4∂u)∂Φ∂v=(∂x1∂v,∂x2∂v,∂x3∂v,∂x4∂v)
and
∂ Φ ∂ w = ( ∂ x 1 ∂ w , ∂ x 2 ∂ w , ∂ x 3 ∂ w , ∂ x 4 ∂ w ) ∂ Φ ∂ w = ∂ x 1 ∂ w , ∂ x 2 ∂ w , ∂ x 3 ∂ w , ∂ x 4 ∂ w (del Phi)/(del w)=((delx^(1))/(del w),(delx^(2))/(del w),(delx^(3))/(del w),(delx^(4))/(del w))\frac{\partial \boldsymbol{\Phi}}{\partial w}=\left(\frac{\partial x^{1}}{\partial w}, \frac{\partial x^{2}}{\partial w}, \frac{\partial x^{3}}{\partial w}, \frac{\partial x^{4}}{\partial w}\right)∂Φ∂w=(∂x1∂w,∂x2∂w,∂x3∂w,∂x4∂w)
We can show a 3 -form ω ω omega\omegaω acting on the tangent vectors ∂ Φ ∂ u , ∂ Φ ∂ v ∂ Φ ∂ u , ∂ Φ ∂ v (del Phi)/(del u),(del Phi)/(del v)\frac{\partial \Phi}{\partial u}, \frac{\partial \Phi}{\partial v}∂Φ∂u,∂Φ∂v and ∂ Φ ∂ w ∂ Φ ∂ w (del Phi)/(del w)\frac{\partial \Phi}{\partial w}∂Φ∂w by
ω ( ∂ Φ ∂ u , ∂ Φ ∂ v , ∂ Φ ∂ w ) ω ∂ Φ ∂ u , ∂ Φ ∂ v , ∂ Φ ∂ w omega((del Phi)/(del u),(del Phi)/(del v),(del Phi)/(del w))\omega\left(\frac{\partial \boldsymbol{\Phi}}{\partial u}, \frac{\partial \boldsymbol{\Phi}}{\partial v}, \frac{\partial \boldsymbol{\Phi}}{\partial w}\right)ω(∂Φ∂u,∂Φ∂v,∂Φ∂w)
Equation ( 2.4 .1 ) ( 2.4 .1 ) (2.4.1)(2.4 .1)(2.4.1)
( ω 1 ∧ ω 2 ∧ ω 3 ) ( v 1 , v 2 , v 3 ) = | ω 1 ( v 1 ) ω 1 ( v 2 ) ω 1 ( v 3 ) ω 2 ( v 1 ) ω 2 ( v 2 ) ω 2 ( v 3 ) ω 3 ( v 1 ) ω 3 ( v 2 ) ω 3 ( v 3 ) | ω 1 ∧ ω 2 ∧ ω 3 v 1 , v 2 , v 3 = ω 1 v 1      ω 1 v 2      ω 1 v 3 ω 2 v 1      ω 2 v 2      ω 2 v 3 ω 3 v 1      ω 3 v 2      ω 3 v 3 (omega_(1)^^omega_(2)^^omega_(3))(v_(1),v_(2),v_(3))=|[omega_(1)(v_(1)),omega_(1)(v_(2)),omega_(1)(v_(3))],[omega_(2)(v_(1)),omega_(2)(v_(2)),omega_(2)(v_(3))],[omega_(3)(v_(1)),omega_(3)(v_(2)),omega_(3)(v_(3))]|\left(\omega_{1} \wedge \omega_{2} \wedge \omega_{3}\right)\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right)=\left|\begin{array}{lll} \omega_{1}\left(\mathbf{v}_{1}\right) & \omega_{1}\left(\mathbf{v}_{2}\right) & \omega_{1}\left(\mathbf{v}_{3}\right) \\ \omega_{2}\left(\mathbf{v}_{1}\right) & \omega_{2}\left(\mathbf{v}_{2}\right) & \omega_{2}\left(\mathbf{v}_{3}\right) \\ \omega_{3}\left(\mathbf{v}_{1}\right) & \omega_{3}\left(\mathbf{v}_{2}\right) & \omega_{3}\left(\mathbf{v}_{3}\right) \end{array}\right|(ω1∧ω2∧ω3)(v1,v2,v3)=|ω1(v1)ω1(v2)ω1(v3)ω2(v1)ω2(v2)ω2(v3)ω3(v1)ω3(v2)ω3(v3)|
tells us how a 3 -form acts on three vectors. If, for example, our 3 -form ω = d x 1 ∧ d x 2 ∧ d x 4 ω = d x 1 ∧ d x 2 ∧ d x 4 omega=dx^(1)^^dx^(2)^^dx^(4)\omega=d x^{1} \wedge d x^{2} \wedge d x^{4}ω=dx1∧dx2∧dx4, we can rewrite this equation in terms of tangent vectors as
( d x 1 ∧ d x 2 ∧ d x 4 ) ( ∂ Φ ∂ u , ∂ Φ ∂ v , ∂ Φ ∂ w ) = | d x 1 ( ∂ Φ ∂ Ψ ) d x 1 ( ∂ Φ ∂ z ) d x 1 ( ∂ Φ ∂ Ψ ) d x 2 ( ∂ Φ ∂ Ψ ) d x 2 ( ∂ Φ ∂ u ) d x 2 ( ∂ Φ ∂ Ψ ) d x 4 ( ∂ Φ ∂ u ) d x 4 ( ∂ Φ ∂ v ) d x 4 ( ∂ Φ ∂ w ) | d x 1 ∧ d x 2 ∧ d x 4 ∂ Φ ∂ u , ∂ Φ ∂ v , ∂ Φ ∂ w = d x 1 ∂ Φ ∂ Ψ      d x 1 ∂ Φ ∂ z      d x 1 ∂ Φ ∂ Ψ d x 2 ∂ Φ ∂ Ψ      d x 2 ∂ Φ ∂ u      d x 2 ∂ Φ ∂ Ψ d x 4 ∂ Φ ∂ u      d x 4 ∂ Φ ∂ v      d x 4 ∂ Φ ∂ w (dx^(1)^^dx^(2)^^dx^(4))((del Phi)/(del u),(del Phi)/(del v),(del Phi)/(del w))=|[dx^(1)((del Phi)/(del Psi)),dx^(1)((del Phi)/(del z)),dx^(1)((del Phi)/(del Psi))],[dx^(2)((del Phi)/(del Psi)),dx^(2)((del Phi)/(del u)),dx^(2)((del Phi)/(del Psi))],[dx^(4)((del Phi)/(del u)),dx^(4)((del Phi)/(del v)),dx^(4)((del Phi)/(del w))]|\left(d x^{1} \wedge d x^{2} \wedge d x^{4}\right)\left(\frac{\partial \boldsymbol{\Phi}}{\partial u}, \frac{\partial \boldsymbol{\Phi}}{\partial v}, \frac{\partial \boldsymbol{\Phi}}{\partial w}\right)=\left|\begin{array}{lll} d x^{1}\left(\frac{\partial \Phi}{\partial \Psi}\right) & d x^{1}\left(\frac{\partial \Phi}{\partial z}\right) & d x^{1}\left(\frac{\partial \Phi}{\partial \Psi}\right) \\ d x^{2}\left(\frac{\partial \Phi}{\partial \Psi}\right) & d x^{2}\left(\frac{\partial \Phi}{\partial u}\right) & d x^{2}\left(\frac{\partial \Phi}{\partial \Psi}\right) \\ d x^{4}\left(\frac{\partial \Phi}{\partial u}\right) & d x^{4}\left(\frac{\partial \Phi}{\partial v}\right) & d x^{4}\left(\frac{\partial \Phi}{\partial w}\right) \end{array}\right|(dx1∧dx2∧dx4)(∂Φ∂u,∂Φ∂v,∂Φ∂w)=|dx1(∂Φ∂Ψ)dx1(∂Φ∂z)dx1(∂Φ∂Ψ)dx2(∂Φ∂Ψ)dx2(∂Φ∂u)dx2(∂Φ∂Ψ)dx4(∂Φ∂u)dx4(∂Φ∂v)dx4(∂Φ∂w)|
(2.4.2) = | ∂ x 1 ∂ u ∂ x 1 ∂ v ∂ x 1 ∂ w ∂ x 2 ∂ u ∂ x 2 ∂ v ∂ x 2 ∂ w ∂ x 4 ∂ u ∂ x 4 ∂ v ∂ x 4 ∂ w | (2.4.2) = ∂ x 1 ∂ u ∂ x 1 ∂ v ∂ x 1 ∂ w ∂ x 2 ∂ u ∂ x 2 ∂ v ∂ x 2 ∂ w ∂ x 4 ∂ u ∂ x 4 ∂ v ∂ x 4 ∂ w {:(2.4.2)=|[(delx^(1))/(del u),(delx^(1))/(del v),(delx^(1))/(del w)],[(delx^(2))/(del u),(delx^(2))/(del v),(delx^(2))/(del w)],[(delx^(4))/(del u),(delx^(4))/(del v),(delx^(4))/(del w)]|:}=\left|\begin{array}{lll} \frac{\partial x^{1}}{\partial u} & \frac{\partial x^{1}}{\partial v} & \frac{\partial x^{1}}{\partial w} \tag{2.4.2}\\ \frac{\partial x^{2}}{\partial u} & \frac{\partial x^{2}}{\partial v} & \frac{\partial x^{2}}{\partial w} \\ \frac{\partial x^{4}}{\partial u} & \frac{\partial x^{4}}{\partial v} & \frac{\partial x^{4}}{\partial w} \end{array}\right|(2.4.2)=|∂x1∂u∂x1∂v∂x1∂w∂x2∂u∂x2∂v∂x2∂w∂x4∂u∂x4∂v∂x4∂w|
Hopefully, you can now see the pattern for differential forms of higher dimension. For example, a k k kkk-form ω = d x 1 ∧ ⋯ ∧ d x k ω = d x 1 ∧ ⋯ ∧ d x k omega=dx^(1)^^cdots^^dx^(k)\omega=d x^{1} \wedge \cdots \wedge d x^{k}ω=dx1∧⋯∧dxk acting on k k kkk tangent vectors ( ∂ Φ ∂ u 1 , … , ∂ Φ ∂ u k ) ∂ Φ ∂ u 1 , … , ∂ Φ ∂ u k ((del Phi)/(delu^(1)),dots,(del Phi)/(delu^(k)))\left(\frac{\partial \Phi}{\partial u^{1}}, \ldots, \frac{\partial \Phi}{\partial u^{k}}\right)(∂Φ∂u1,…,∂Φ∂uk) in R k R k R^(k)\mathbb{R}^{k}Rk would be given by
( d x 1 ∧ ⋯ ∧ d x k ) ( ∂ Φ ∂ u 1 , … , ∂ Φ ∂ u k ) = | ∂ x 1 ∂ u 1 ⋯ ∂ x 1 ∂ u k ⋮ ⋮ ∂ x k ∂ u 1 ⋯ ∂ x k ∂ u k | d x 1 ∧ ⋯ ∧ d x k ∂ Φ ∂ u 1 , … , ∂ Φ ∂ u k = ∂ x 1 ∂ u 1 ⋯ ∂ x 1 ∂ u k ⋮ ⋮ ∂ x k ∂ u 1 ⋯ ∂ x k ∂ u k (dx^(1)^^cdots^^dx^(k))((del Phi)/(delu^(1)),dots,(del Phi)/(delu^(k)))=|[(delx^(1))/(delu^(1)),cdots,(delx^(1))/(delu^(k))],[vdots,,vdots],[(delx^(k))/(delu^(1)),cdots,(delx^(k))/(delu^(k))]|\left(d x^{1} \wedge \cdots \wedge d x^{k}\right)\left(\frac{\partial \boldsymbol{\Phi}}{\partial u^{1}}, \ldots, \frac{\partial \boldsymbol{\Phi}}{\partial u^{k}}\right)=\left|\begin{array}{ccc} \frac{\partial x^{1}}{\partial u^{1}} & \cdots & \frac{\partial x^{1}}{\partial u^{k}} \\ \vdots & & \vdots \\ \frac{\partial x^{k}}{\partial u^{1}} & \cdots & \frac{\partial x^{k}}{\partial u^{k}} \end{array}\right|(dx1∧⋯∧dxk)(∂Φ∂u1,…,∂Φ∂uk)=|∂x1∂u1⋯∂x1∂uk⋮⋮∂xk∂u1⋯∂xk∂uk|
And so on.

3 Converting between differential forms and vectors

In R 3 R 3 R^(3)\mathbb{R}^{3}R3 differential forms nicely correspond with various scalar and vector fields. We'll be using these correspondences in subsequent chapters to derive some useful vector calculus formulas.

3.10 -forms

We've already noted that a smooth function is a 0 -form. So a scalar field, such as ϕ ( x , y , z ) ϕ ( x , y , z ) phi(x,y,z)\phi(x, y, z)ϕ(x,y,z), is also a 0 -form.

3.2 1-forms

In R 3 R 3 R^(3)\mathbb{R}^{3}R3, in general coordinates, every vector field can be associated with a corresponding 1 -form and a corresponding 2 -form. How does that work? For non-Cartesian coordinates we need something called a metric tensor g i j g i j g_(ij)g_{i j}gij, a type of function that measures infinitesimal distances on a manifold. The metric tensor encodes information about both the curvature of the manifold and the chosen coordinate system, and also allows us to convert vector components to 1 -form components and vice versa.
However, in Cartesian coordinates, things are much simpler; the metric tensor is effectively just a collection of ones and zeros and the correspondences couldn't be more straightforward.
For 1-forms:
(3.2.1) d x ⇔ e ^ x , d y ⇔ e ^ y , d z ⇔ e ^ z (3.2.1) d x ⇔ e ^ x , d y ⇔ e ^ y , d z ⇔ e ^ z {:(3.2.1)dx<=> hat(e)_(x)","dy<=> hat(e)_(y)","dz<=> hat(e)_(z):}\begin{equation*} d x \Leftrightarrow \hat{\mathbf{e}}_{x}, d y \Leftrightarrow \hat{\mathbf{e}}_{y}, d z \Leftrightarrow \hat{\mathbf{e}}_{z} \tag{3.2.1} \end{equation*}(3.2.1)dx⇔e^x,dy⇔e^y,dz⇔e^z
So a vector field
w ( x , y , z ) = f 1 e ^ x + f 2 e ^ y + f 3 e ^ z w ( x , y , z ) = f 1 e ^ x + f 2 e ^ y + f 3 e ^ z w(x,y,z)=f_(1) hat(e)_(x)+f_(2) hat(e)_(y)+f_(3) hat(e)_(z)\mathbf{w}(x, y, z)=f_{1} \hat{\mathbf{e}}_{x}+f_{2} \hat{\mathbf{e}}_{y}+f_{3} \hat{\mathbf{e}}_{z}w(x,y,z)=f1e^x+f2e^y+f3e^z
is associated with the 1 -form
(3.2.2) ω 1 = f 1 d x + f 2 d y + f 3 d z (3.2.2) ω 1 = f 1 d x + f 2 d y + f 3 d z {:(3.2.2)omega_(1)=f_(1)dx+f_(2)dy+f_(3)dz:}\begin{equation*} \omega_{1}=f_{1} d x+f_{2} d y+f_{3} d z \tag{3.2.2} \end{equation*}(3.2.2)ω1=f1dx+f2dy+f3dz
In other words, we can regard the basis 1 -forms as equivalent to the standard basis vectors, as illustrated in Figure 3.1.
Figure 3.1: Basis 1-forms in Cartesian coordinates.

3.3 2-forms

Figure 3.2: Basis 2-forms in Cartesian coordinates.
For 2-forms:
(3.3.1) d y ∧ d z ⇔ e ^ x , d z ∧ d x ⇔ e ^ y , d x ∧ d y ⇔ e ^ z (3.3.1) d y ∧ d z ⇔ e ^ x , d z ∧ d x ⇔ e ^ y , d x ∧ d y ⇔ e ^ z {:(3.3.1)dy^^dz<=> hat(e)_(x)","dz^^dx<=> hat(e)_(y)","dx^^dy<=> hat(e)_(z):}\begin{equation*} d y \wedge d z \Leftrightarrow \hat{\mathbf{e}}_{x}, d z \wedge d x \Leftrightarrow \hat{\mathbf{e}}_{y}, d x \wedge d y \Leftrightarrow \hat{\mathbf{e}}_{z} \tag{3.3.1} \end{equation*}(3.3.1)dy∧dz⇔e^x,dz∧dx⇔e^y,dx∧dy⇔e^z
So a vector field
w ( x , y , z ) = f 1 e ^ x + f 2 e ^ y + f 3 e ^ z w ( x , y , z ) = f 1 e ^ x + f 2 e ^ y + f 3 e ^ z w(x,y,z)=f_(1) hat(e)_(x)+f_(2) hat(e)_(y)+f_(3) hat(e)_(z)\mathbf{w}(x, y, z)=f_{1} \hat{\mathbf{e}}_{x}+f_{2} \hat{\mathbf{e}}_{y}+f_{3} \hat{\mathbf{e}}_{z}w(x,y,z)=f1e^x+f2e^y+f3e^z
is also associated with the 2 -form
(3.3.2) ω 2 = f 1 d y ∧ d z + f 2 d z ∧ d x + f 3 d x ∧ d y (3.3.2) ω 2 = f 1 d y ∧ d z + f 2 d z ∧ d x + f 3 d x ∧ d y {:(3.3.2)omega_(2)=f_(1)dy^^dz+f_(2)dz^^dx+f_(3)dx^^dy:}\begin{equation*} \omega_{2}=f_{1} d y \wedge d z+f_{2} d z \wedge d x+f_{3} d x \wedge d y \tag{3.3.2} \end{equation*}(3.3.2)ω2=f1dy∧dz+f2dz∧dx+f3dx∧dy
The basis 2-forms d y ∧ d z , d z ∧ d x d y ∧ d z , d z ∧ d x dy^^dz,dz^^dxd y \wedge d z, d z \wedge d xdy∧dz,dz∧dx and d x ∧ d y d x ∧ d y dx^^dyd x \wedge d ydx∧dy are equivalent to the standard unit basis vectors, but in a surfacey sort of way, as indicated in Figure 3.2. We can regard d y ∧ d z , d z ∧ d x d y ∧ d z , d z ∧ d x dy^^dz,dz^^dxd y \wedge d z, d z \wedge d xdy∧dz,dz∧dx and d x ∧ d y d x ∧ d y dx^^dyd x \wedge d ydx∧dy as flux-calculators that tell us how much fluid or field flows through surfaces aligned with the coordinate axes. To justify this flux interpretation of a 2 -form, consider two vectors, u u u\mathbf{u}u and v v v\mathbf{v}v, that span an oriented parallelogram P P PPP. Equation (2.3.4) tells us
ω 2 ( u , v ) = | f 1 u 1 v 1 f 2 u 2 v 2 f 3 u 3 v 3 | = w ⋅ ( u × v ) ω 2 ( u , v ) = f 1 u 1 v 1 f 2 u 2 v 2 f 3 u 3 v 3 = w ⋅ ( u × v ) omega_(2)(u,v)=|[f_(1),u^(1),v^(1)],[f_(2),u^(2),v^(2)],[f_(3),u^(3),v^(3)]|=w*(uxxv)\omega_{2}(\mathbf{u}, \mathbf{v})=\left|\begin{array}{ccc} f_{1} & u^{1} & v^{1} \\ f_{2} & u^{2} & v^{2} \\ f_{3} & u^{3} & v^{3} \end{array}\right|=\mathbf{w} \cdot(\mathbf{u} \times \mathbf{v})ω2(u,v)=|f1u1v1f2u2v2f3u3v3|=w⋅(u×v)
If we let n n n\mathbf{n}n be a unit vector normal to P P PPP (and in the same direction as u × v u × v uxxv\mathbf{u} \times \mathbf{v}u×v ), we can write
ω 2 ( u , v ) = | f 1 u 1 v 1 f 2 u 2 v 2 f 3 u 3 v 3 | = ( w ⋅ n ) ( n ⋅ ( u × v ) ) = ( w ⋅ n ) ‖ u × v ‖ ω 2 ( u , v ) = f 1 u 1 v 1 f 2 u 2 v 2 f 3 u 3 v 3 = ( w ⋅ n ) ( n ⋅ ( u × v ) ) = ( w ⋅ n ) ‖ u × v ‖ {:[omega_(2)(u","v)=|[f_(1),u^(1),v^(1)],[f_(2),u^(2),v^(2)],[f_(3),u^(3),v^(3)]|=(w*n)(n*(uxxv))],[=(w*n)||uxxv||]:}\begin{gathered} \omega_{2}(\mathbf{u}, \mathbf{v})=\left|\begin{array}{ccc} f_{1} & u^{1} & v^{1} \\ f_{2} & u^{2} & v^{2} \\ f_{3} & u^{3} & v^{3} \end{array}\right|=(\mathbf{w} \cdot \mathbf{n})(\mathbf{n} \cdot(\mathbf{u} \times \mathbf{v})) \\ =(\mathbf{w} \cdot \mathbf{n})\|\mathbf{u} \times \mathbf{v}\| \end{gathered}ω2(u,v)=|f1u1v1f2u2v2f3u3v3|=(w⋅n)(n⋅(u×v))=(w⋅n)‖u×v‖
(where ‖ u × v ‖ ‖ u × v ‖ ||uxxv||\|\mathbf{u} \times \mathbf{v}\|‖u×v‖ is the magnitude of the cross product u × v u × v uxxv\mathbf{u} \times \mathbf{v}u×v )
(3.3.3) = ( w ⋅ n ) × area of P (3.3.3) = ( w ⋅ n ) ×  area of  P {:(3.3.3)=(w*n)xx" area of "P:}\begin{equation*} =(\mathbf{w} \cdot \mathbf{n}) \times \text { area of } P \tag{3.3.3} \end{equation*}(3.3.3)=(w⋅n)× area of P
which tells us the component of w w w\mathbf{w}w in the normal direction passing through P P PPP. If w represents a constant fluid flow, then we just need (3.3.3) to find the amount of fluid passing through the parallelogram P P PPP per unit time, ie the flux. Otherwise, if w w w\mathbf{w}w describes a varying fluid flow, we need to use the surface integral version of (3.3.3), as described in chapter 8 .

3.4 3-forms

In chapter 5 we'll see that the 3 -form (5.3.3)
(3.4.1) ( ∂ f 1 ∂ x + ∂ f 2 ∂ y + ∂ f 3 ∂ z ) d x ∧ d y ∧ d z (3.4.1) ∂ f 1 ∂ x + ∂ f 2 ∂ y + ∂ f 3 ∂ z d x ∧ d y ∧ d z {:(3.4.1)((delf_(1))/(del x)+(delf_(2))/(del y)+(delf_(3))/(del z))dx^^dy^^dz:}\begin{equation*} \left(\frac{\partial f_{1}}{\partial x}+\frac{\partial f_{2}}{\partial y}+\frac{\partial f_{3}}{\partial z}\right) d x \wedge d y \wedge d z \tag{3.4.1} \end{equation*}(3.4.1)(∂f1∂x+∂f2∂y+∂f3∂z)dx∧dy∧dz
is associated with the divergence of a vector field, which is the scalar field
ϕ 1 ( x , y , z ) = ( ∂ f 1 ∂ x + ∂ f 2 ∂ y + ∂ f 3 ∂ z ) ϕ 1 ( x , y , z ) = ∂ f 1 ∂ x + ∂ f 2 ∂ y + ∂ f 3 ∂ z phi_(1)(x,y,z)=((delf_(1))/(del x)+(delf_(2))/(del y)+(delf_(3))/(del z))\phi_{1}(x, y, z)=\left(\frac{\partial f_{1}}{\partial x}+\frac{\partial f_{2}}{\partial y}+\frac{\partial f_{3}}{\partial z}\right)ϕ1(x,y,z)=(∂f1∂x+∂f2∂y+∂f3∂z)
We started this chapter by noting that a scalar field is a 0 -form. We have now come full circle by seeing that, in R 3 R 3 R^(3)\mathbb{R}^{3}R3, the 3 -form (3.4.1) is also associated with a scalar
field. Table 3.1 summarises the correspondence between differential forms and vector calculus in R 3 R 3 R^(3)\mathbb{R}^{3}R3.
Table 3.1: Correspondence between differential forms and scalar/vector fields.

4 Differentiation

We need to be able to differentiate differential forms. We do this using something called the exterior derivative, which generalises the derivative of smooth functions (aka 0 -forms) and incorporates the quirky, sign-flipping behaviour of the wedge product. The exterior derivative operator d d ddd changes a k k kkk-form to a ( k + 1 ) ( k + 1 ) (k+1)(k+1)(k+1)-form according to these four rules:
  1. d d ddd is a linear operator, ie for the differential k k kkk-forms ω ω omega\omegaω and ν ν nu\nuν
d ( a ω + b ν ) = a d ω + b d ν d ( a ω + b ν ) = a d ω + b d ν d(a omega+b nu)=ad omega+bd nud(a \omega+b \nu)=a d \omega+b d \nud(aω+bν)=adω+bdν
where a a aaa and b b bbb are real numbers.
  1. The exterior derivative of a 0 -form (ie a smooth function) is the differential (2.2.1) that we've already met. So, for a function f f fff in a space M M MMM with coordinates x 1 , … , x n x 1 , … , x n x^(1),dots,x^(n)x^{1}, \ldots, x^{n}x1,…,xn
d f = ∂ f ∂ x 1 d x 1 + ∂ f ∂ x 2 d x 2 + ⋯ + ∂ f ∂ x n d x n d f = ∂ f ∂ x 1 d x 1 + ∂ f ∂ x 2 d x 2 + ⋯ + ∂ f ∂ x n d x n df=(del f)/(delx^(1))dx^(1)+(del f)/(delx^(2))dx^(2)+cdots+(del f)/(delx^(n))dx^(n)d f=\frac{\partial f}{\partial x^{1}} d x^{1}+\frac{\partial f}{\partial x^{2}} d x^{2}+\cdots+\frac{\partial f}{\partial x^{n}} d x^{n}df=∂f∂x1dx1+∂f∂x2dx2+⋯+∂f∂xndxn
  1. If ω ω omega\omegaω is a p p ppp-form and ν ν nu\nuν is a q q qqq-form, then (this corresponds to the product rule of ordinary calculus)
d ( ω ∧ ν ) = d ω ∧ ν + ( − 1 ) p ω ∧ d ν d ( ω ∧ ν ) = d ω ∧ ν + ( − 1 ) p ω ∧ d ν d(omega^^nu)=d omega^^nu+(-1)^(p)omega^^d nud(\omega \wedge \nu)=d \omega \wedge \nu+(-1)^{p} \omega \wedge d \nud(ω∧ν)=dω∧ν+(−1)pω∧dν
  1. d ( d ω ) = 0 d ( d ω ) = 0 d(d omega)=0d(d \omega)=0d(dω)=0. Or, more succinctly, d d = 0 d d = 0 dd=0d d=0dd=0.
A few examples in R 3 R 3 R^(3)\mathbb{R}^{3}R3 should make the process clearer.
Example 4.1. For a function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x, y, z)f(x,y,z), find the exterior derivative of the 1-form d f d f dfd fdf.
Rule 4 , d ( d ω ) = 0 4 , d ( d ω ) = 0 4,d(d omega)=04, d(d \omega)=04,d(dω)=0 (in this case d ( d f ) = 0 d ( d f ) = 0 d(df)=0d(d f)=0d(df)=0 ) tells us that the answer must be zero. However, it's instructive to see how that result inexorably emerges out of the calculation.
The derivative of f f fff is given by rule 2 , ie
d f = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z d f = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z df=(del f)/(del x)dx+(del f)/(del y)dy+(del f)/(del z)dzd f=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y+\frac{\partial f}{\partial z} d zdf=∂f∂xdx+∂f∂ydy+∂f∂zdz
which is, of course, a 1-form. Taking the external derivative gives
d ( d f ) = d ( ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z ) d ( d f ) = d ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z d(df)=d((del f)/(del x)dx+(del f)/(del y)dy+(del f)/(del z)dz)d(d f)=d\left(\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y+\frac{\partial f}{\partial z} d z\right)d(df)=d(∂f∂xdx+∂f∂ydy+∂f∂zdz)
= ∂ 2 f ∂ x ∂ x d x ∧ d x + ∂ 2 f ∂ y ∂ x d y ∧ d x + ∂ 2 f ∂ z ∂ x d z ∧ d x + ∂ 2 f ∂ x ∂ y d x ∧ d y + ∂ 2 f ∂ y ∂ y d y ∧ d y + ∂ 2 f ∂ z ∂ y d z ∧ d y + ∂ 2 f ∂ x ∂ z d x ∧ d z + ∂ 2 f ∂ y ∂ z d y ∧ d z + ∂ 2 f ∂ z ∂ z d z ∧ d z = ∂ 2 f ∂ y ∂ x d y ∧ d x + ∂ 2 f ∂ z ∂ x d z ∧ d x + ∂ 2 f ∂ x ∂ y d x ∧ d y + ∂ 2 f ∂ z ∂ y d z ∧ d y + ∂ 2 f ∂ x ∂ z d x ∧ d z + ∂ 2 f ∂ y ∂ z d y ∧ d z = ∂ 2 f ∂ x ∂ x d x ∧ d x + ∂ 2 f ∂ y ∂ x d y ∧ d x + ∂ 2 f ∂ z ∂ x d z ∧ d x + ∂ 2 f ∂ x ∂ y d x ∧ d y + ∂ 2 f ∂ y ∂ y d y ∧ d y + ∂ 2 f ∂ z ∂ y d z ∧ d y + ∂ 2 f ∂ x ∂ z d x ∧ d z + ∂ 2 f ∂ y ∂ z d y ∧ d z + ∂ 2 f ∂ z ∂ z d z ∧ d z = ∂ 2 f ∂ y ∂ x d y ∧ d x + ∂ 2 f ∂ z ∂ x d z ∧ d x + ∂ 2 f ∂ x ∂ y d x ∧ d y + ∂ 2 f ∂ z ∂ y d z ∧ d y + ∂ 2 f ∂ x ∂ z d x ∧ d z + ∂ 2 f ∂ y ∂ z d y ∧ d z {:[=(del^(2)f)/(del x del x)dx^^dx+(del^(2)f)/(del y del x)dy^^dx+(del^(2)f)/(del z del x)dz^^dx],[+(del^(2)f)/(del x del y)dx^^dy+(del^(2)f)/(del y del y)dy^^dy+(del^(2)f)/(del z del y)dz^^dy],[+(del^(2)f)/(del x del z)dx^^dz+(del^(2)f)/(del y del z)dy^^dz+(del^(2)f)/(del z del z)dz^^dz],[=(del^(2)f)/(del y del x)dy^^dx+(del^(2)f)/(del z del x)dz^^dx],[+(del^(2)f)/(del x del y)dx^^dy+(del^(2)f)/(del z del y)dz^^dy],[quad+(del^(2)f)/(del x del z)dx^^dz+(del^(2)f)/(del y del z)dy^^dz]:}\begin{aligned} &=\frac{\partial^{2} f}{\partial x \partial x} d x \wedge d x+\frac{\partial^{2} f}{\partial y \partial x} d y \wedge d x+\frac{\partial^{2} f}{\partial z \partial x} d z \wedge d x \\ &+\frac{\partial^{2} f}{\partial x \partial y} d x \wedge d y+\frac{\partial^{2} f}{\partial y \partial y} d y \wedge d y+\frac{\partial^{2} f}{\partial z \partial y} d z \wedge d y \\ &+\frac{\partial^{2} f}{\partial x \partial z} d x \wedge d z+\frac{\partial^{2} f}{\partial y \partial z} d y \wedge d z+\frac{\partial^{2} f}{\partial z \partial z} d z \wedge d z \\ &=\frac{\partial^{2} f}{\partial y \partial x} d y \wedge d x+\frac{\partial^{2} f}{\partial z \partial x} d z \wedge d x \\ &+\frac{\partial^{2} f}{\partial x \partial y} d x \wedge d y+\frac{\partial^{2} f}{\partial z \partial y} d z \wedge d y \\ & \quad+\frac{\partial^{2} f}{\partial x \partial z} d x \wedge d z+\frac{\partial^{2} f}{\partial y \partial z} d y \wedge d z \end{aligned}=∂2f∂x∂xdx∧dx+∂2f∂y∂xdy∧dx+∂2f∂z∂xdz∧dx+∂2f∂x∂ydx∧dy+∂2f∂y∂ydy∧dy+∂2f∂z∂ydz∧dy+∂2f∂x∂zdx∧dz+∂2f∂y∂zdy∧dz+∂2f∂z∂zdz∧dz=∂2f∂y∂xdy∧dx+∂2f∂z∂xdz∧dx+∂2f∂x∂ydx∧dy+∂2f∂z∂ydz∧dy+∂2f∂x∂zdx∧dz+∂2f∂y∂zdy∧dz
Which we can rewrite, with the differentials in the correct orders, as
d ( d f ) = ( ∂ 2 f ∂ y ∂ z − ∂ 2 f ∂ z ∂ y ) d y ∧ d z + ( ∂ 2 f ∂ z ∂ x − ∂ 2 f ∂ x ∂ z ) d z ∧ d x + ( ∂ 2 f ∂ x ∂ y − ∂ 2 f ∂ y ∂ x ) d x ∧ d y d ( d f ) = ∂ 2 f ∂ y ∂ z − ∂ 2 f ∂ z ∂ y d y ∧ d z + ∂ 2 f ∂ z ∂ x − ∂ 2 f ∂ x ∂ z d z ∧ d x + ∂ 2 f ∂ x ∂ y − ∂ 2 f ∂ y ∂ x d x ∧ d y {:[d(df)=((del^(2)f)/(del y del z)-(del^(2)f)/(del z del y))dy^^dz],[+((del^(2)f)/(del z del x)-(del^(2)f)/(del x del z))dz^^dx],[+((del^(2)f)/(del x del y)-(del^(2)f)/(del y del x))dx^^dy]:}\begin{aligned} & d(d f)=\left(\frac{\partial^{2} f}{\partial y \partial z}-\frac{\partial^{2} f}{\partial z \partial y}\right) d y \wedge d z \\ &+\left(\frac{\partial^{2} f}{\partial z \partial x}-\frac{\partial^{2} f}{\partial x \partial z}\right) d z \wedge d x \\ &+\left(\frac{\partial^{2} f}{\partial x \partial y}-\frac{\partial^{2} f}{\partial y \partial x}\right) d x \wedge d y \end{aligned}d(df)=(∂2f∂y∂z−∂2f∂z∂y)dy∧dz+(∂2f∂z∂x−∂2f∂x∂z)dz∧dx+(∂2f∂x∂y−∂2f∂y∂x)dx∧dy
But partial derivatives commutes, ie ∂ 2 f ∂ x ∂ y = ∂ 2 f ∂ y ∂ x ∂ 2 f ∂ x ∂ y = ∂ 2 f ∂ y ∂ x (del^(2)f)/(del x del y)=(del^(2)f)/(del y del x)\frac{\partial^{2} f}{\partial x \partial y}=\frac{\partial^{2} f}{\partial y \partial x}∂2f∂x∂y=∂2f∂y∂x. Therefore
( ∂ 2 f ∂ y ∂ z − ∂ 2 f ∂ z ∂ y ) = ( ∂ 2 f ∂ z ∂ x − ∂ 2 f ∂ x ∂ z ) = ( ∂ 2 f ∂ x ∂ y − ∂ 2 f ∂ y ∂ x ) = 0 ∂ 2 f ∂ y ∂ z − ∂ 2 f ∂ z ∂ y = ∂ 2 f ∂ z ∂ x − ∂ 2 f ∂ x ∂ z = ∂ 2 f ∂ x ∂ y − ∂ 2 f ∂ y ∂ x = 0 ((del^(2)f)/(del y del z)-(del^(2)f)/(del z del y))=((del^(2)f)/(del z del x)-(del^(2)f)/(del x del z))=((del^(2)f)/(del x del y)-(del^(2)f)/(del y del x))=0\left(\frac{\partial^{2} f}{\partial y \partial z}-\frac{\partial^{2} f}{\partial z \partial y}\right)=\left(\frac{\partial^{2} f}{\partial z \partial x}-\frac{\partial^{2} f}{\partial x \partial z}\right)=\left(\frac{\partial^{2} f}{\partial x \partial y}-\frac{\partial^{2} f}{\partial y \partial x}\right)=0(∂2f∂y∂z−∂2f∂z∂y)=(∂2f∂z∂x−∂2f∂x∂z)=(∂2f∂x∂y−∂2f∂y∂x)=0
and we find
d ( d f ) = 0 d ( d f ) = 0 d(df)=0d(d f)=0d(df)=0
which is what rule 4 tells us.
In passing, it's worth noting that a differential form ω ω omega\omegaω is said to be closed if d ω = 0 d ω = 0 d omega=0d \omega=0dω=0.
Example 4.2. (a) Find d ( f d x ) d ( f d x ) d(fdx)d(f d x)d(fdx) for a function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x, y, z)f(x,y,z). (b) Find d ( f d y ) d ( f d y ) d(fdy)d(f d y)d(fdy) for f ( x , y , z ) = x 3 y 2 z 4 f ( x , y , z ) = x 3 y 2 z 4 f(x,y,z)=x^(3)y^(2)z^(4)f(x, y, z)=x^{3} y^{2} z^{4}f(x,y,z)=x3y2z4; in other words, find the exterior derivative of the 1-form x 3 y 2 z 4 d y x 3 y 2 z 4 d y x^(3)y^(2)z^(4)dyx^{3} y^{2} z^{4} d yx3y2z4dy.
(a) Recall from section 2.3.1 that the convention is not to use the ∧ ∧ ^^\wedge∧ when multiplying a form by a function, ie to write f d x f d x fdxf d xfdx and not f ∧ d x f ∧ d x f^^dxf \wedge d xf∧dx. A function is a 0 -form, so using rule 3 (with p = 0 p = 0 p=0p=0p=0 ) we get
d ( f ∧ d x ) = d ( f d x ) = d f ∧ d x + ( − 1 ) 0 f ∧ d ( d x ) d ( f ∧ d x ) = d ( f d x ) = d f ∧ d x + ( − 1 ) 0 f ∧ d ( d x ) d(f^^dx)=d(fdx)=df^^dx+(-1)^(0)f^^d(dx)d(f \wedge d x)=d(f d x)=d f \wedge d x+(-1)^{0} f \wedge d(d x)d(f∧dx)=d(fdx)=df∧dx+(−1)0f∧d(dx)
= d f ∧ d x + 1 × f ∧ 0 = d f ∧ d x = d f ∧ d x + 1 × f ∧ 0 = d f ∧ d x {:[=df^^dx+1xx f^^0],[=df^^dx]:}\begin{gathered} =d f \wedge d x+1 \times f \wedge 0 \\ =d f \wedge d x \end{gathered}=df∧dx+1×f∧0=df∧dx
where we've also used rule 4 , d ( d x ) = 0 4 , d ( d x ) = 0 4,d(dx)=04, d(d x)=04,d(dx)=0.
(b) Using this result we can write
d ( f d y ) = d f ∧ d y d ( f d y ) = d f ∧ d y d(fdy)=df^^dyd(f d y)=d f \wedge d yd(fdy)=df∧dy
which gives
d ( x 3 y 2 z 4 d y ) = ( 3 x 2 y 2 z 4 d x + 2 x 3 y z 4 d y + 4 x 3 y 2 z 3 d z ) ∧ d y = 3 x 2 y 2 z 4 d x ∧ d y − 4 x 3 y 2 z 3 d y ∧ d z d x 3 y 2 z 4 d y = 3 x 2 y 2 z 4 d x + 2 x 3 y z 4 d y + 4 x 3 y 2 z 3 d z ∧ d y = 3 x 2 y 2 z 4 d x ∧ d y − 4 x 3 y 2 z 3 d y ∧ d z {:[d(x^(3)y^(2)z^(4)dy)=(3x^(2)y^(2)z^(4)dx+2x^(3)yz^(4)dy+4x^(3)y^(2)z^(3)dz)^^dy],[=3x^(2)y^(2)z^(4)dx^^dy-4x^(3)y^(2)z^(3)dy^^dz]:}\begin{gathered} d\left(x^{3} y^{2} z^{4} d y\right)=\left(3 x^{2} y^{2} z^{4} d x+2 x^{3} y z^{4} d y+4 x^{3} y^{2} z^{3} d z\right) \wedge d y \\ =3 x^{2} y^{2} z^{4} d x \wedge d y-4 x^{3} y^{2} z^{3} d y \wedge d z \end{gathered}d(x3y2z4dy)=(3x2y2z4dx+2x3yz4dy+4x3y2z3dz)∧dy=3x2y2z4dx∧dy−4x3y2z3dy∧dz
Note that 2 x 3 y z 4 d y ∧ d y = 0 2 x 3 y z 4 d y ∧ d y = 0 2x^(3)yz^(4)dy^^dy=02 x^{3} y z^{4} d y \wedge d y=02x3yz4dy∧dy=0 because d y ∧ d y = 0 d y ∧ d y = 0 dy^^dy=0d y \wedge d y=0dy∧dy=0. And we've changed the sign of the 4 x 3 y 2 z 3 d z ∧ d y 4 x 3 y 2 z 3 d z ∧ d y 4x^(3)y^(2)z^(3)dz^^dy4 x^{3} y^{2} z^{3} d z \wedge d y4x3y2z3dz∧dy term to get the differentials d y , d z d y , d z dy,dzd y, d zdy,dz in the correct order. Note also that we start with a 1 -form and finish with a 2 -form.
Example 4.3. From Bryan [7]. Find d η d η d etad \etadη for the 2 -form η = ( x + z 2 ) d x ∧ d y η = x + z 2 d x ∧ d y eta=(x+z^(2))dx^^dy\eta=\left(x+z^{2}\right) d x \wedge d yη=(x+z2)dx∧dy.
We use rule 3 with ω = ( x + z 2 ) d x ω = x + z 2 d x omega=(x+z^(2))dx\omega=\left(x+z^{2}\right) d xω=(x+z2)dx and ν = d y ν = d y nu=dy\nu=d yν=dy. As ω ω omega\omegaω is a 1 -form, p = 1 p = 1 p=1p=1p=1 and we get
d η = d ( ( x + z 2 ) d x ) ∧ d y + ( − 1 ) 1 ( x + z 2 ) d x ∧ d ( d y ) = ( d x + 2 z d z ) d x ∧ d y − ( x + z 2 ) d x × 0 = 2 z d z ∧ d x ∧ d y = − 2 z d x ∧ d z ∧ d y = 2 z d x ∧ d y ∧ d z d η = d x + z 2 d x ∧ d y + ( − 1 ) 1 x + z 2 d x ∧ d ( d y ) = ( d x + 2 z d z ) d x ∧ d y − x + z 2 d x × 0 = 2 z d z ∧ d x ∧ d y = − 2 z d x ∧ d z ∧ d y = 2 z d x ∧ d y ∧ d z {:[d eta=d((x+z^(2))dx)^^dy+(-1)^(1)(x+z^(2))dx^^d(dy)],[=(dx+2zdz)dx^^dy-(x+z^(2))dx xx0],[=2zdz^^dx^^dy],[=-2zdx^^dz^^dy],[=2zdx^^dy^^dz]:}\begin{aligned} & d \eta=d\left(\left(x+z^{2}\right) d x\right) \wedge d y+(-1)^{1}\left(x+z^{2}\right) d x \wedge d(d y) \\ &=(d x+2 z d z) d x \wedge d y-\left(x+z^{2}\right) d x \times 0 \\ &=2 z d z \wedge d x \wedge d y \\ &=-2 z d x \wedge d z \wedge d y \\ &=2 z d x \wedge d y \wedge d z \end{aligned}dη=d((x+z2)dx)∧dy+(−1)1(x+z2)dx∧d(dy)=(dx+2zdz)dx∧dy−(x+z2)dx×0=2zdz∧dx∧dy=−2zdx∧dz∧dy=2zdx∧dy∧dz
Note that we start with a 2 -form and finish with a 3 -form.
When looking at the wedge product in section 2.3.1 we saw that, because of (2.3.1)
d x i ∧ d x i = 0 d x i ∧ d x i = 0 dx^(i)^^dx^(i)=0d x^{i} \wedge d x^{i}=0dxi∧dxi=0
there are no 4 -forms or higher in R 3 R 3 R^(3)\mathbb{R}^{3}R3. So what happens if we try to sneak in a 4 -form by taking the exterior derivative of a 3 -form in R 3 R 3 R^(3)\mathbb{R}^{3}R3 ?
Example 4.4. Find d ( f d x ∧ d y ∧ d z ) d ( f d x ∧ d y ∧ d z ) d(fdx^^dy^^dz)d(f d x \wedge d y \wedge d z)d(fdx∧dy∧dz) for a function f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x, y, z)f(x,y,z).
We use rule 3 with ω = f d x ω = f d x omega=fdx\omega=f d xω=fdx and ν = d y ∧ d z . ω ν = d y ∧ d z . ω nu=dy^^dz.omega\nu=d y \wedge d z . \omegaν=dy∧dz.ω is a 1 -form, so p = 1 p = 1 p=1p=1p=1, and we get
d ( f d x ∧ d y ∧ d z ) = d ( f d x ) ∧ d y ∧ d z + ( − 1 ) 1 ( f d x ) ∧ d ( d y ∧ d z ) = d f ∧ d x ∧ d y ∧ d z = ( ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z ) ∧ d x ∧ d y ∧ d z = 0 d ( f d x ∧ d y ∧ d z ) = d ( f d x ) ∧ d y ∧ d z + ( − 1 ) 1 ( f d x ) ∧ d ( d y ∧ d z ) = d f ∧ d x ∧ d y ∧ d z = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z ∧ d x ∧ d y ∧ d z = 0 {:[d(fdx^^dy^^dz)=d(fdx)^^dy^^dz+(-1)^(1)(fdx)^^d(dy^^dz)],[=df^^dx^^dy^^dz],[=((del f)/(del x)dx+(del f)/(del y)dy+(del f)/(del z)dz)^^dx^^dy^^dz],[=0]:}\begin{aligned} & d(f d x \wedge d y \wedge d z)=d(f d x) \wedge d y \wedge d z+(-1)^{1}(f d x) \wedge d(d y \wedge d z) \\ & =d f \wedge d x \wedge d y \wedge d z \\ & =\left(\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y+\frac{\partial f}{\partial z} d z\right) \wedge d x \wedge d y \wedge d z \\ & =0 \end{aligned}d(fdx∧dy∧dz)=d(fdx)∧dy∧dz+(−1)1(fdx)∧d(dy∧dz)=df∧dx∧dy∧dz=(∂f∂xdx+∂f∂ydy+∂f∂zdz)∧dx∧dy∧dz=0
because d x ∧ d x = d y ∧ d y = d z ∧ d z = 0 d x ∧ d x = d y ∧ d y = d z ∧ d z = 0 dx^^dx=dy^^dy=dz^^dz=0d x \wedge d x=d y \wedge d y=d z \wedge d z=0dx∧dx=dy∧dy=dz∧dz=0. Which is what we'd expect - in R 3 R 3 R^(3)\mathbb{R}^{3}R3 the exterior derivative of a 3 -form is zero.

5 Div, grad and curl

There are four kinds of differential forms in R 3 R 3 R^(3)\mathbb{R}^{3}R3 : 0-forms, 1-forms, 2 -forms and 3-forms. We've just seen that in R 3 R 3 R^(3)\mathbb{R}^{3}R3 the exterior derivative of a 3 -form vanishes. By taking the exterior derivative of 0 -forms, 1 -forms and 2 -forms we can express the three important operators of vector calculus - grad, curl and div - in the language of differential forms.

5.1 Gradient and 0 -forms

The gradient (written grad f grad f grad f\operatorname{grad} fgradf or ∇ f ∇ f grad f\nabla f∇f ) of a scalar field f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x, y, z)f(x,y,z) is a vector field that tells us two things. First, at any point, the gradient vector points in the direction of greatest rate of increase of f f fff. Second, the magnitude of the gradient vector gives the rate of increase of f f fff in that direction. The gradient of a scalar field f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x, y, z)f(x,y,z) is given by
(5.1.1) grad f = ∇ f = ∂ f ∂ x e ^ x + ∂ f ∂ y e ^ y + ∂ f ∂ z e ^ z (5.1.1) grad f = ∇ f = ∂ f ∂ x e ^ x + ∂ f ∂ y e ^ y + ∂ f ∂ z e ^ z {:(5.1.1)grad f=grad f=(del f)/(del x) hat(e)_(x)+(del f)/(del y) hat(e)_(y)+(del f)/(del z) hat(e)_(z):}\begin{equation*} \operatorname{grad} f=\nabla f=\frac{\partial f}{\partial x} \hat{\mathbf{e}}_{x}+\frac{\partial f}{\partial y} \hat{\mathbf{e}}_{y}+\frac{\partial f}{\partial z} \hat{\mathbf{e}}_{z} \tag{5.1.1} \end{equation*}(5.1.1)gradf=∇f=∂f∂xe^x+∂f∂ye^y+∂f∂ze^z
Using the correspondence (3.2.1)
d x ⇔ e ^ x , d y ⇔ e ^ y , d z ⇔ e ^ z d x ⇔ e ^ x , d y ⇔ e ^ y , d z ⇔ e ^ z dx<=> hat(e)_(x),dy<=> hat(e)_(y),dz<=> hat(e)_(z)d x \Leftrightarrow \hat{\mathbf{e}}_{x}, d y \Leftrightarrow \hat{\mathbf{e}}_{y}, d z \Leftrightarrow \hat{\mathbf{e}}_{z}dx⇔e^x,dy⇔e^y,dz⇔e^z
this gradient vector field can be associated with the exterior derivative of the 0 -form f ( x , y , z ) f ( x , y , z ) f(x,y,z)f(x, y, z)f(x,y,z), which is the 1 -form
d f = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z d f = ∂ f ∂ x d x + ∂ f ∂ y d y + ∂ f ∂ z d z df=(del f)/(del x)dx+(del f)/(del y)dy+(del f)/(del z)dzd f=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y+\frac{\partial f}{\partial z} d zdf=∂f∂xdx+∂f∂ydy+∂f∂zdz

5.2 Curl and 1 -forms

The curl (written curl v v v\mathbf{v}v or ∇ × v ∇ × v grad xxv\nabla \times \mathbf{v}∇×v ) of a vector field v v v\mathbf{v}v is itself a vector and is a measure of the field's rotation at a point. Let the vector field represent the flow of some fluid. Now imagine inserting a tiny sphere into the fluid. The sphere is ingeniously secured in such a way that it isn't carried along with the fluid's current but is free to rotate in any direction. The rotation of the sphere is a measure of the curl of the vector field; no rotation, means zero curl. The direction of the vector curl v v v\mathbf{v}v is along the sphere's axis of rotation, as determined by the right-hand rule. Curl the fingers of your right hand in the direction of the sphere's rotation with your thumb perpendicular to your fist. Your thumb is then pointing in the direction of curl v v v\mathbf{v}v.
The curl of a vector field v ( x , y , z ) = f 1 e ^ x + f 2 e ^ y + f 3 e ^ z v ( x , y , z ) = f 1 e ^ x + f 2 e ^ y + f 3 e ^ z v(x,y,z)=f_(1) hat(e)_(x)+f_(2) hat(e)_(y)+f_(3) hat(e)_(z)\mathbf{v}(x, y, z)=f_{1} \hat{\mathbf{e}}_{x}+f_{2} \hat{\mathbf{e}}_{y}+f_{3} \hat{\mathbf{e}}_{z}v(x,y,z)=f1e^x+f2e^y+f3e^z is given by
(5.2.1) curl v = ∇ × v = ( ∂ f 3 ∂ y − ∂ f 2 ∂ z ) e ^ x + ( ∂ f 1 ∂ z − ∂ f 3 ∂ x ) e ^ y + ( ∂ f 2 ∂ x − ∂ f 1 ∂ y ) e ^ z (5.2.1) curl v = ∇ × v = ∂ f 3 ∂ y − ∂ f 2 ∂ z e ^ x + ∂ f 1 ∂ z − ∂ f 3 ∂ x e ^ y + ∂ f 2 ∂ x − ∂ f 1 ∂ y e ^ z {:(5.2.1)curlv=grad xxv=((delf_(3))/(del y)-(delf_(2))/(del z)) hat(e)_(x)+((delf_(1))/(del z)-(delf_(3))/(del x)) hat(e)_(y)+((delf_(2))/(del x)-(delf_(1))/(del y)) hat(e)_(z):}\begin{equation*} \operatorname{curl} \mathbf{v}=\nabla \times \mathbf{v}=\left(\frac{\partial f_{3}}{\partial y}-\frac{\partial f_{2}}{\partial z}\right) \hat{\mathbf{e}}_{x}+\left(\frac{\partial f_{1}}{\partial z}-\frac{\partial f_{3}}{\partial x}\right) \hat{\mathbf{e}}_{y}+\left(\frac{\partial f_{2}}{\partial x}-\frac{\partial f_{1}}{\partial y}\right) \hat{\mathbf{e}}_{z} \tag{5.2.1} \end{equation*}(5.2.1)curlv=∇×v=(∂f3∂y−∂f2∂z)e^x+(∂f1∂z−∂f3∂x)e^y+(∂f2∂x−∂f1∂y)e^z
or, in determinant form,
curl v = | e ^ x e ^ y e ^ z ∂ ∂ x ∂ ∂ y ∂ ∂ z f 1 f 2 f 3 | curl v = e ^ x e ^ y e ^ z ∂ ∂ x ∂ ∂ y ∂ ∂ z f 1 f 2 f 3 curlv=|[ hat(e)_(x), hat(e)_(y), hat(e)_(z)],[(del)/(del x),(del)/(del y),(del)/(del z)],[f_(1),f_(2),f_(3)]|\operatorname{curl} \mathbf{v}=\left|\begin{array}{ccc} \hat{\mathbf{e}}_{x} & \hat{\mathbf{e}}_{y} & \hat{\mathbf{e}}_{\boldsymbol{z}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f_{1} & f_{2} & f_{3} \end{array}\right|curlv=|e^xe^ye^z∂∂x∂∂y∂∂zf1f2f3|
Now let's take the exterior derivative of the 1 -form (3.2.2) associated with v ( x , y , z ) v ( x , y , z ) v(x,y,z)\mathbf{v}(x, y, z)v(x,y,z) :
ω 1 = f 1 d x + f 2 d y + f 3 d z ω 1 = f 1 d x + f 2 d y + f 3 d z omega_(1)=f_(1)dx+f_(2)dy+f_(3)dz\omega_{1}=f_{1} d x+f_{2} d y+f_{3} d zω1=f1dx+f2dy+f3dz
to get
(5.2.2) d ω 1 = d ( f 1 d x + f 2 d y + f 3 d z ) = d f 1 ∧ d x + d f 2 ∧ d y + d f 3 ∧ d z = ( ∂ f 1 ∂ x d x + ∂ f 1 ∂ y d y + ∂ f 1 ∂ z d z ) ∧ d x + ( ∂ f 2 ∂ x d x + ∂ f 2 ∂ y d y + ∂ f 2 ∂ z d z ) ∧ d y + ( ∂ f 3 ∂ x d x + ∂ f 3 ∂ y d y + ∂ f 3 ∂ z d z ) ∧ d z = ∂ f 1 ∂ y d y ∧ d x + ∂ f 1 ∂ z d z ∧ d x + ∂ f 2 ∂ x d x ∧ d y + ∂ f 2 ∂ z d z ∧ d y + ∂ f 3 ∂ x d x ∧ d z + ∂ f 3 ∂ y d y ∧ d z = ( ∂ f 3 ∂ y − ∂ f 2 ∂ z ) d y ∧ d z + ( ∂ f 1 ∂ z − ∂ f 3 ∂ x ) d z ∧ d x + ( ∂ f 2 ∂ x − ∂ f 1 ∂ y ) d x ∧ d y (5.2.2) d ω 1 = d f 1 d x + f 2 d y + f 3 d z = d f 1 ∧ d x + d f 2 ∧ d y + d f 3 ∧ d z = ∂ f 1 ∂ x d x + ∂ f 1 ∂ y d y + ∂ f 1 ∂ z d z ∧ d x + ∂ f 2 ∂ x d x + ∂ f 2 ∂ y d y + ∂ f 2 ∂ z d z ∧ d y + ∂ f 3 ∂ x d x + ∂ f 3 ∂ y d y + ∂ f 3 ∂ z d z ∧ d z = ∂ f 1 ∂ y d y ∧ d x + ∂ f 1 ∂ z d z ∧ d x + ∂ f 2 ∂ x d x ∧ d y + ∂ f 2 ∂ z d z ∧ d y + ∂ f 3 ∂ x d x ∧ d z + ∂ f 3 ∂ y d y ∧ d z = ∂ f 3 ∂ y − ∂ f 2 ∂ z d y ∧ d z + ∂ f 1 ∂ z − ∂ f 3 ∂ x d z ∧ d x + ∂ f 2 ∂ x − ∂ f 1 ∂ y d x ∧ d y {:[(5.2.2)domega_(1)=d(f_(1)dx+f_(2)dy+f_(3)dz)],[=df_(1)^^dx+df_(2)^^dy+df_(3)^^dz],[=((delf_(1))/(del x)dx+(delf_(1))/(del y)dy+(delf_(1))/(del z)dz)^^dx],[+((delf_(2))/(del x)dx+(delf_(2))/(del y)dy+(delf_(2))/(del z)dz)^^dy],[+((delf_(3))/(del x)dx+(delf_(3))/(del y)dy+(delf_(3))/(del z)dz)^^dz],[=(delf_(1))/(del y)dy^^dx+(delf_(1))/(del z)dz^^dx],[+(delf_(2))/(del x)dx^^dy+(delf_(2))/(del z)dz^^dy],[+(delf_(3))/(del x)dx^^dz+(delf_(3))/(del y)dy^^dz],[=((delf_(3))/(del y)-(delf_(2))/(del z))dy^^dz+((delf_(1))/(del z)-(delf_(3))/(del x))dz^^dx+((delf_(2))/(del x)-(delf_(1))/(del y))dx^^dy]:}\begin{gather*} d \omega_{1}=d\left(f_{1} d x+f_{2} d y+f_{3} d z\right) \tag{5.2.2}\\ =d f_{1} \wedge d x+d f_{2} \wedge d y+d f_{3} \wedge d z \\ =\left(\frac{\partial f_{1}}{\partial x} d x+\frac{\partial f_{1}}{\partial y} d y+\frac{\partial f_{1}}{\partial z} d z\right) \wedge d x \\ +\left(\frac{\partial f_{2}}{\partial x} d x+\frac{\partial f_{2}}{\partial y} d y+\frac{\partial f_{2}}{\partial z} d z\right) \wedge d y \\ +\left(\frac{\partial f_{3}}{\partial x} d x+\frac{\partial f_{3}}{\partial y} d y+\frac{\partial f_{3}}{\partial z} d z\right) \wedge d z \\ =\frac{\partial f_{1}}{\partial y} d y \wedge d x+\frac{\partial f_{1}}{\partial z} d z \wedge d x \\ +\frac{\partial f_{2}}{\partial x} d x \wedge d y+\frac{\partial f_{2}}{\partial z} d z \wedge d y \\ +\frac{\partial f_{3}}{\partial x} d x \wedge d z+\frac{\partial f_{3}}{\partial y} d y \wedge d z \\ =\left(\frac{\partial f_{3}}{\partial y}-\frac{\partial f_{2}}{\partial z}\right) d y \wedge d z+\left(\frac{\partial f_{1}}{\partial z}-\frac{\partial f_{3}}{\partial x}\right) d z \wedge d x+\left(\frac{\partial f_{2}}{\partial x}-\frac{\partial f_{1}}{\partial y}\right) d x \wedge d y \end{gather*}(5.2.2)dω1=d(f1dx+f2dy+f3dz)=df1∧dx+df2∧dy+df3∧dz=(∂f1∂xdx+∂f1∂ydy+∂f1∂zdz)∧dx+(∂f2∂xdx+∂f2∂ydy+∂f2∂zdz)∧dy+(∂f3∂xdx+∂f3∂ydy+∂f3∂zdz)∧dz=∂f1∂ydy∧dx+∂f1∂zdz∧dx+∂f2∂xdx∧dy+∂f2∂zdz∧dy+∂f3∂xdx∧dz+∂f3∂ydy∧dz=(∂f3∂y−∂f2∂z)dy∧dz+(∂f1∂z−∂f3∂x)dz∧dx+(∂f2∂x−∂f1∂y)dx∧dy
where we've changed the signs to get the differentials in the correct order and used d x i ∧ d x i = 0 d x i ∧ d x i = 0 dx^(i)^^dx^(i)=0d x^{i} \wedge d x^{i}=0dxi∧dxi=0
If we use the correspondence ( 3.3 .1 ) ( 3.3 .1 ) (3.3.1)(3.3 .1)(3.3.1)
d y ∧ d z ⇔ e ^ x , d z ∧ d x ⇔ e ^ y , d x ∧ d y ⇔ e ^ z d y ∧ d z ⇔ e ^ x , d z ∧ d x ⇔ e ^ y , d x ∧ d y ⇔ e ^ z dy^^dz<=> hat(e)_(x),dz^^dx<=> hat(e)_(y),dx^^dy<=> hat(e)_(z)d y \wedge d z \Leftrightarrow \hat{\mathbf{e}}_{x}, d z \wedge d x \Leftrightarrow \hat{\mathbf{e}}_{y}, d x \wedge d y \Leftrightarrow \hat{\mathbf{e}}_{z}dy∧dz⇔e^x,dz∧dx⇔e^y,dx∧dy⇔e^z
we can nicely associate the exterior derivative d ω 1 d ω 1 domega_(1)d \omega_{1}dω1 with the equation for curl (5.2.1).

5.3 Divergence and 2-forms

The divergence (written div w div w divw\operatorname{div} \mathbf{w}divw or ∇ ⋅ w ∇ ⋅ w grad*w\nabla \cdot \mathbf{w}∇⋅w ) of a vector field w w w\mathbf{w}w is a scalar field and is a measure of the net flow of fluid through an infinitesimally small closed surface surrounding a point. If more fluid is leaving than entering the surface, the point is a source (think of a running tap as a source of water) and the divergence is positive. If more fluid is entering than leaving the surface, the point is a sink (think of water flowing down a drain) and the divergence is negative. If the same amount of fluid is entering as leaving, there is zero divergence.
The divergence of a vector field w ( x , y , z ) = f 1 e ^ x + f 2 e ^ y + f 3 e ^ z w ( x , y , z ) = f 1 e ^ x + f 2 e ^ y + f 3 e ^ z w(x,y,z)=f_(1) hat(e)_(x)+f_(2) hat(e)_(y)+f_(3) hat(e)_(z)\mathbf{w}(x, y, z)=f_{1} \hat{\mathbf{e}}_{x}+f_{2} \hat{\mathbf{e}}_{y}+f_{3} \hat{\mathbf{e}}_{z}w(x,y,z)=f1e^x+f2e^y+f3e^z is given by
(5.3.1) div w = ∇ ⋅ w = ∂ f 1 ∂ x + ∂ f 2 ∂ y + ∂ f 3 ∂ z (5.3.1) div w = ∇ ⋅ w = ∂ f 1 ∂ x + ∂ f 2 ∂ y + ∂ f 3 ∂ z {:(5.3.1)divw=grad*w=(delf_(1))/(del x)+(delf_(2))/(del y)+(delf_(3))/(del z):}\begin{equation*} \operatorname{div} \mathbf{w}=\nabla \cdot \mathbf{w}=\frac{\partial f_{1}}{\partial x}+\frac{\partial f_{2}}{\partial y}+\frac{\partial f_{3}}{\partial z} \tag{5.3.1} \end{equation*}(5.3.1)divw=∇⋅w=∂f1∂x+∂f2∂y+∂f3∂z
If we take the exterior derivative of the 2 -form (3.3.2) associated with w ( x , y , z ) w ( x , y , z ) w(x,y,z)\mathbf{w}(x, y, z)w(x,y,z) :
ω 2 = f 1 d y ∧ d z + f 2 d z ∧ d x + f 3 d x ∧ d y ω 2 = f 1 d y ∧ d z + f 2 d z ∧ d x + f 3 d x ∧ d y omega_(2)=f_(1)dy^^dz+f_(2)dz^^dx+f_(3)dx^^dy\omega_{2}=f_{1} d y \wedge d z+f_{2} d z \wedge d x+f_{3} d x \wedge d yω2=f1dy∧dz+f2dz∧dx+f3dx∧dy
we get
(5.3.2) d ω 2 = d ( f 1 d y ∧ d z + f 2 d z ∧ d x + f 3 d x ∧ d y ) = d f 1 ∧ d y ∧ d z + d f 2 ∧ d z ∧ d x + d f 3 ∧ d x ∧ d y = ∂ f 1 ∂ x d x ∧ d y ∧ d z + ∂ f 2 ∂ y d y ∧ d z ∧ d x + ∂ f 3 ∂ z d z ∧ d x ∧ d y (5.3.3) = ( ∂ f 1 ∂ x + ∂ f 2 ∂ y + ∂ f 3 ∂ z ) d x ∧ d y ∧ d z (5.3.2) d ω 2 = d f 1 d y ∧ d z + f 2 d z ∧ d x + f 3 d x ∧ d y = d f 1 ∧ d y ∧ d z + d f 2 ∧ d z ∧ d x + d f 3 ∧ d x ∧ d y = ∂ f 1 ∂ x d x ∧ d y ∧ d z + ∂ f 2 ∂ y d y ∧ d z ∧ d x + ∂ f 3 ∂ z d z ∧ d x ∧ d y (5.3.3) = ∂ f 1 ∂ x + ∂ f 2 ∂ y + ∂ f 3 ∂ z d x ∧ d y ∧ d z {:[(5.3.2)domega_(2)=d(f_(1)dy^^dz+f_(2)dz^^dx+f_(3)dx^^dy)],[=df_(1)^^dy^^dz+df_(2)^^dz^^dx+df_(3)^^dx^^dy],[=(delf_(1))/(del x)dx^^dy^^dz+(delf_(2))/(del y)dy^^dz^^dx+(delf_(3))/(del z)dz^^dx^^dy],[(5.3.3)=((delf_(1))/(del x)+(delf_(2))/(del y)+(delf_(3))/(del z))dx^^dy^^dz]:}\begin{align*} & d \omega_{2}=d\left(f_{1} d y \wedge d z+f_{2} d z \wedge d x+f_{3} d x \wedge d y\right) \tag{5.3.2}\\ & =d f_{1} \wedge d y \wedge d z+d f_{2} \wedge d z \wedge d x+d f_{3} \wedge d x \wedge d y \\ & =\frac{\partial f_{1}}{\partial x} d x \wedge d y \wedge d z+\frac{\partial f_{2}}{\partial y} d y \wedge d z \wedge d x+\frac{\partial f_{3}}{\partial z} d z \wedge d x \wedge d y \\ & =\left(\frac{\partial f_{1}}{\partial x}+\frac{\partial f_{2}}{\partial y}+\frac{\partial f_{3}}{\partial z}\right) d x \wedge d y \wedge d z \tag{5.3.3} \end{align*}(5.3.2)dω2=d(f1dy∧dz+f2dz∧dx+f3dx∧dy)=df1∧dy∧dz+df2∧dz∧dx+df3∧dx∧dy=∂f1∂xdx∧dy∧dz+∂f2∂ydy∧dz∧dx+∂f3∂zdz∧dx∧dy(5.3.3)=(∂f1∂x+∂f2∂y+∂f3∂z)dx∧dy∧dz
The coefficient of this 3 -form is the scalar field
( ∂ f 1 ∂ x + ∂ f 2 ∂ y + ∂ f 3 ∂ z ) ∂ f 1 ∂ x + ∂ f 2 ∂ y + ∂ f 3 ∂ z ((delf_(1))/(del x)+(delf_(2))/(del y)+(delf_(3))/(del z))\left(\frac{\partial f_{1}}{\partial x}+\frac{\partial f_{2}}{\partial y}+\frac{\partial f_{3}}{\partial z}\right)(∂f1∂x+∂f2∂y+∂f3∂z)
which is the equation for divergence (5.3.1). So we can associate the exterior derivative d ω 2 d ω 2 domega_(2)d \omega_{2}dω2 with the divergence ∇ ⋅ v ∇ ⋅ v grad*v\nabla \cdot \mathbf{v}∇⋅v.
Table 5.1 summarises the relationships between the vector calculus operators grad, curl and div and their associated differential forms.
Table 5.1: Grad, curl and div with vector calculus and differential forms.

5.4 A couple of vector identities

In chapter 4 we saw that rule 4 for taking the exterior derivative is d ( d ω ) = 0 d ( d ω ) = 0 d(d omega)=0d(d \omega)=0d(dω)=0. We can use this rule to easily obtain two important vector calculus identities.
First, recall that if we take the exterior derivative of the 1-form
ω 1 = f 1 d x + f 2 d y + f 3 d z ω 1 = f 1 d x + f 2 d y + f 3 d z omega_(1)=f_(1)dx+f_(2)dy+f_(3)dz\omega_{1}=f_{1} d x+f_{2} d y+f_{3} d zω1=f1dx+f2dy+f3dz
associated with the vector field v ( x , y , z ) v ( x , y , z ) v(x,y,z)\mathbf{v}(x, y, z)v(x,y,z), we get ( 5.2 .2 ) ( 5.2 .2 ) (5.2.2)(5.2 .2)(5.2.2)
d ω 1 = d ( f 1 d x + f 2 d y + f 3 d z ) d ω 1 = d f 1 d x + f 2 d y + f 3 d z domega_(1)=d(f_(1)dx+f_(2)dy+f_(3)dz)d \omega_{1}=d\left(f_{1} d x+f_{2} d y+f_{3} d z\right)dω1=d(f1dx+f2dy+f3dz)
which corresponds to curl v = ∇ × v v = ∇ × v v=grad xxv\mathbf{v}=\nabla \times \mathbf{v}v=∇×v. Now substitute for ω 1 ω 1 omega_(1)\omega_{1}ω1 the 1-form d f d f dfd fdf, which can be associated with the gradient vector field
grad f = ∇ f = ∂ f ∂ x e ^ x + ∂ f ∂ y e ^ y + ∂ f ∂ z e ^ z grad f = ∇ f = ∂ f ∂ x e ^ x + ∂ f ∂ y e ^ y + ∂ f ∂ z e ^ z grad f=grad f=(del f)/(del x) hat(e)_(x)+(del f)/(del y) hat(e)_(y)+(del f)/(del z) hat(e)_(z)\operatorname{grad} f=\nabla f=\frac{\partial f}{\partial x} \hat{\mathbf{e}}_{x}+\frac{\partial f}{\partial y} \hat{\mathbf{e}}_{y}+\frac{\partial f}{\partial z} \hat{\mathbf{e}}_{z}gradf=∇f=∂f∂xe^x+∂f∂ye^y+∂f∂ze^z
But d ( d f ) = 0 d ( d f ) = 0 d(df)=0d(d f)=0d(df)=0.
  • This tells us that
∇ × ∇ f = 0 ∇ × ∇ f = 0 grad xx grad f=0\nabla \times \nabla f=0∇×∇f=0
ie the curl of the gradient is zero.
Second, recall that if we take the exterior derivative of the 2 -form
ω 2 = f 1 d y ∧ d z + f 2 d z ∧ d x + f 3 d x ∧ d y ω 2 = f 1 d y ∧ d z + f 2 d z ∧ d x + f 3 d x ∧ d y omega_(2)=f_(1)dy^^dz+f_(2)dz^^dx+f_(3)dx^^dy\omega_{2}=f_{1} d y \wedge d z+f_{2} d z \wedge d x+f_{3} d x \wedge d yω2=f1dy∧dz+f2dz∧dx+f3dx∧dy
associated with the vector field w ( x , y , z ) w ( x , y , z ) w(x,y,z)\mathbf{w}(x, y, z)w(x,y,z), we get (5.3.2)
d ω 2 = d ( f 1 d y ∧ d z + f 2 d z ∧ d x + f 3 d x ∧ d y ) d ω 2 = d f 1 d y ∧ d z + f 2 d z ∧ d x + f 3 d x ∧ d y domega_(2)=d(f_(1)dy^^dz+f_(2)dz^^dx+f_(3)dx^^dy)d \omega_{2}=d\left(f_{1} d y \wedge d z+f_{2} d z \wedge d x+f_{3} d x \wedge d y\right)dω2=d(f1dy∧dz+f2dz∧dx+f3dx∧dy)
the coefficient of which,
( ∂ f 1 ∂ x + ∂ f 2 ∂ y + ∂ f 3 ∂ z ) ∂ f 1 ∂ x + ∂ f 2 ∂ y + ∂ f 3 ∂ z ((delf_(1))/(del x)+(delf_(2))/(del y)+(delf_(3))/(del z))\left(\frac{\partial f_{1}}{\partial x}+\frac{\partial f_{2}}{\partial y}+\frac{\partial f_{3}}{\partial z}\right)(∂f1∂x+∂f2∂y+∂f3∂z)
corresponds to the divergence ∇ ⋅ w ∇ ⋅ w grad*w\nabla \cdot \mathbf{w}∇⋅w. Now substitute for ω 2 ω 2 omega_(2)\omega_{2}ω2 the 2 -form d ω 1 d ω 1 domega_(1)d \omega_{1}dω1, which can be associated with the curl vector field
curl v = ∇ × v = ( ∂ f 3 ∂ y − ∂ f 2 ∂ z ) e ^ x + ( ∂ f 1 ∂ z − ∂ f 3 ∂ x ) e ^ y + ( ∂ f 2 ∂ x − ∂ f 1 ∂ y ) e ^ z curl v = ∇ × v = ∂ f 3 ∂ y − ∂ f 2 ∂ z e ^ x + ∂ f 1 ∂ z − ∂ f 3 ∂ x e ^ y + ∂ f 2 ∂ x − ∂ f 1 ∂ y e ^ z curlv=grad xxv=((delf_(3))/(del y)-(delf_(2))/(del z)) hat(e)_(x)+((delf_(1))/(del z)-(delf_(3))/(del x)) hat(e)_(y)+((delf_(2))/(del x)-(delf_(1))/(del y)) hat(e)_(z)\operatorname{curl} \mathbf{v}=\nabla \times \mathbf{v}=\left(\frac{\partial f_{3}}{\partial y}-\frac{\partial f_{2}}{\partial z}\right) \hat{\mathbf{e}}_{x}+\left(\frac{\partial f_{1}}{\partial z}-\frac{\partial f_{3}}{\partial x}\right) \hat{\mathbf{e}}_{y}+\left(\frac{\partial f_{2}}{\partial x}-\frac{\partial f_{1}}{\partial y}\right) \hat{\mathbf{e}}_{z}curlv=∇×v=(∂f3∂y−∂f2∂z)e^x+(∂f1∂z−∂f3∂x)e^y+(∂f2∂x−∂f1∂y)e^z
But d ( d ω 1 ) = 0 d d ω 1 = 0 d(domega_(1))=0d\left(d \omega_{1}\right)=0d(dω1)=0.
  • This tells us that
∇ ⋅ ( ∇ × v ) = 0 ∇ ⋅ ( ∇ × v ) = 0 grad*(grad xxv)=0\nabla \cdot(\nabla \times \mathbf{v})=0∇⋅(∇×v)=0
ie the divergence of the curl is zero.

6 Orientation

We'll see in the next chapter that we integrate differential k k kkk-forms over oriented k k kkk manifolds (a 1-form over a one-dimensional curve, for example), so we need to look at the notion of orientation. A manifold that is orientable has one of two orientations. We can think of orientation as one of two non-equivalent ways in which objects are situated in space. Intuitively, a change of orientation describes the difference between a page of text and its mirror image; each being a reflection of the other. We could arbitrarily label these two orientations 'ordinary' and 'mirror'. No matter how much we might rotate or stretch the page of text, we won't be able to change its orientation from 'ordinary' to 'mirror'. Only a reflection can do that.
  • The key practical point when doing calculations is that if we muddle up the orientation our answer may end up with the wrong sign.
We'll begin by considering the orientation of a vector space, namely the basis vectors of R n R n R^(n)\mathbb{R}^{n}Rn. First, geometrically, for R 1 , R 2 R 1 , R 2 R^(1),R^(2)\mathbb{R}^{1}, \mathbb{R}^{2}R1,R2 and R 3 R 3 R^(3)\mathbb{R}^{3}R3, then progressing to a more general algebraic analysis for R n R n R^(n)\mathbb{R}^{n}Rn.
positive orientation
negative orientation
Figure 6.1: Oriented bases for R 1 , R 2 R 1 , R 2 R^(1),R^(2)\mathbb{R}^{1}, \mathbb{R}^{2}R1,R2 and R 3 R 3 R^(3)\mathbb{R}^{3}R3.
A geometric interpretation of basis orientation is shown in Figure 6.1, which depicts positive and negative oriented bases for R 1 , R 2 R 1 , R 2 R^(1),R^(2)\mathbb{R}^{1}, \mathbb{R}^{2}R1,R2 and R 3 R 3 R^(3)\mathbb{R}^{3}R3. (Note that labelling an orientation 'positive' or 'negative' is completely arbitrary, a matter of convention, not
mathematical fact.) The top row shows the standard or preferred orientations. In R 1 R 1 R^(1)\mathbb{R}^{1}R1 the standard (positive) orientation is given by the basis vector e 1 e 1 e_(1)\mathbf{e}_{1}e1 pointing to the right (ie in the direction of increasing x x xxx ). In R 2 R 2 R^(2)\mathbb{R}^{2}R2 the standard (positive or counterclockwise) orientation is given by noting that the shortest path from the first basis vector e 1 e 1 e_(1)\mathbf{e}_{1}e1 to the second basis vector e 2 e 2 e_(2)\mathbf{e}_{2}e2 is in a counterclockwise direction. In R 3 R 3 R^(3)\mathbb{R}^{3}R3 the standard (positive or right-handed) orientation is given using the right-hand rule, which states that if you hold your right hand so that the fingers curl from e 1 e 1 e_(1)\mathbf{e}_{1}e1 to e 2 e 2 e_(2)\mathbf{e}_{2}e2, your thumb will be pointing in the direction of e 3 e 3 e_(3)\mathbf{e}_{3}e3. (Alternatively, imagine looking down the e 3 e 3 e_(3)\mathbf{e}_{3}e3 vector; the shortest path from e 1 e 1 e_(1)\mathbf{e}_{1}e1 to e 2 e 2 e_(2)\mathbf{e}_{2}e2 is then in a counterclockwise direction.)
The bottom row shows the corresponding opposite orientations. In R 1 R 1 R^(1)\mathbb{R}^{1}R1 the negative orientation is given by the basis vector − e 1 − e 1 -e_(1)-\mathbf{e}_{1}−e1 pointing to the left (ie in the direction of decreasing x ) x ) x)x)x). In R 2 R 2 R^(2)\mathbb{R}^{2}R2 we now make e 2 e 2 e_(2)\mathbf{e}_{2}e2 the first basis vector. The (negative or clockwise) orientation is given by noting that the shortest path from e 2 e 2 e_(2)\mathbf{e}_{2}e2 to e 1 e 1 e_(1)\mathbf{e}_{1}e1 is in a clockwise direction. In R 3 R 3 R^(3)\mathbb{R}^{3}R3 we also now make e 2 e 2 e_(2)\mathbf{e}_{2}e2 the first basis vector and e 1 e 1 e_(1)\mathbf{e}_{1}e1 the second basis vector. The (negative or left-handed) orientation is given using the left-hand rule, which states that if you hold your left hand so that the fingers curl from e 2 e 2 e_(2)\mathbf{e}_{2}e2 to e 1 e 1 e_(1)\mathbf{e}_{1}e1, your thumb will be pointing in the direction of e 3 e 3 e_(3)\mathbf{e}_{3}e3. (Alternatively, imagine looking down the e 3 e 3 e_(3)\mathbf{e}_{3}e3 vector; the shortest path from e 2 e 2 e_(2)\mathbf{e}_{2}e2 to e 1 e 1 e_(1)\mathbf{e}_{1}e1 is then in a clockwise direction.)
Descriptions such as 'to the right', 'to the left', 'counterclockwise', 'clockwise', 'righthanded' and 'left-handed' are somewhat vague. What would counterclockwise mean, for example, if we were trying to understand the orientation of R n R n R^(n)\mathbb{R}^{n}Rn where n > 3 n > 3 n > 3n>3n>3 ? To put things on a precise, mathematical footing, we introduce an algebraic interpretation of orientation. We do this by specifying an orientation for R n R n R^(n)\mathbb{R}^{n}Rn using an ordered basis ( e 1 , … , e n ) e 1 , … , e n (e_(1),dots,e_(n))\left(\mathbf{e}_{1}, \ldots, \mathbf{e}_{n}\right)(e1,…,en). We've already implicitly made use of ordered bases when we referred to rules for finding the shortest path (counterclockwise or clockwise) from the first to the second basis vectors. In effect, we were saying that for a standard positive orientation, the ordered bases for R 2 R 2 R^(2)\mathbb{R}^{2}R2 and R 3 R 3 R^(3)\mathbb{R}^{3}R3 are, respectively, ( e 1 , e 2 ) e 1 , e 2 (e_(1),e_(2))\left(\mathbf{e}_{1}, \mathbf{e}_{2}\right)(e1,e2) and ( e 1 , e 2 , e 3 ) e 1 , e 2 , e 3 (e_(1),e_(2),e_(3))\left(\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\right)(e1,e2,e3). And, for a negative orientation, the ordered bases for R 2 R 2 R^(2)\mathbb{R}^{2}R2 and R 3 R 3 R^(3)\mathbb{R}^{3}R3 are ( e 2 , e 1 ) e 2 , e 1 (e_(2),e_(1))\left(\mathbf{e}_{2}, \mathbf{e}_{1}\right)(e2,e1) and ( e 2 , e 1 , e 3 ) e 2 , e 1 , e 3 (e_(2),e_(1),e_(3))\left(\mathbf{e}_{2}, \mathbf{e}_{1}, \mathbf{e}_{3}\right)(e2,e1,e3). We saw that if we swapped any two vectors in an ordered basis the orientation flips. For example, the orientation in R 2 R 2 R^(2)\mathbb{R}^{2}R2 changes from positive to negative if we change the ordered basis from ( e 1 , e 2 ) e 1 , e 2 (e_(1),e_(2))\left(\mathbf{e}_{1}, \mathbf{e}_{2}\right)(e1,e2) to ( e 2 , e 1 ) e 2 , e 1 (e_(2),e_(1))\left(\mathbf{e}_{2}, \mathbf{e}_{1}\right)(e2,e1). You should be able to see that the orientation also flips if we change the sign of any basis vector. So, referring to the top row, middle diagram in Figure 6.1, if we changed the sign of e 1 e 1 e_(1)\mathbf{e}_{1}e1, the vector would be pointing from the origin to the left, the shortest path from e 1 e 1 e_(1)\mathbf{e}_{1}e1 to e 2 e 2 e_(2)\mathbf{e}_{2}e2 would be in a clockwise direction and the orientation would change from positive to negative.
Objects that are (a) assembled from vectors, (b) change sign if any two vectors are swapped, and (c) change sign if any one vector is multiplied by -1 may well sound familiar. These, of course, are properties of the determinant. In fact, determinants are the key to deciding whether two ordered bases in R n R n R^(n)\mathbb{R}^{n}Rn have the same orientation, which they do if their determinants have the same sign. We can see what this means by assuming the vectors in Figure 6.1 are orthonormal unit basis vectors. We then have:
det ( e 1 ) = | 1 | = 1 , det ( − e 1 ) = | − 1 | = − 1 det e 1 = | 1 | = 1 , det − e 1 = | − 1 | = − 1 det(e_(1))=|1|=1,det(-e_(1))=|-1|=-1\operatorname{det}\left(\mathbf{e}_{1}\right)=|1|=1, \operatorname{det}\left(-\mathbf{e}_{1}\right)=|-1|=-1det(e1)=|1|=1,det(−e1)=|−1|=−1
det ( e 1 , e 2 ) = | 1 0 0 1 | = 1 , det ( e 2 , e 1 ) = | 0 1 1 0 | = − 1 det ( e 1 , e 2 , e 3 ) = | 1 0 0 0 1 0 0 0 1 | = 1 , det ( e 2 , e 1 , e 3 ) = | 0 1 0 1 0 0 0 0 1 | = − 1 det e 1 , e 2 = 1 0 0 1 = 1 , det e 2 , e 1 = 0 1 1 0 = − 1 det e 1 , e 2 , e 3 = 1 0 0 0 1 0 0 0 1 = 1 , det e 2 , e 1 , e 3 = 0 1 0 1 0 0 0 0 1 = − 1 {:[det(e_(1),e_(2))=|[1,0],[0,1]|=1","det(e_(2),e_(1))=|[0,1],[1,0]|=-1],[det(e_(1),e_(2),e_(3))=|[1,0,0],[0,1,0],[0,0,1]|=1","det(e_(2),e_(1),e_(3))=|[0,1,0],[1,0,0],[0,0,1]|=-1]:}\begin{gathered} \operatorname{det}\left(\mathbf{e}_{1}, \mathbf{e}_{2}\right)=\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|=1, \operatorname{det}\left(\mathbf{e}_{2}, \mathbf{e}_{1}\right)=\left|\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right|=-1 \\ \operatorname{det}\left(\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\right)=\left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|=1, \operatorname{det}\left(\mathbf{e}_{2}, \mathbf{e}_{1}, \mathbf{e}_{3}\right)=\left|\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right|=-1 \end{gathered}det(e1,e2)=|1001|=1,det(e2,e1)=|0110|=−1det(e1,e2,e3)=|100010001|=1,det(e2,e1,e3)=|010100001|=−1
The positively oriented bases on the left have positive determinants. The negatively oriented bases on the right have negative determinants. And this analysis also works for non-orthonormal ordered bases in R n R n R^(n)\mathbb{R}^{n}Rn. So if we have two ordered bases ( u 1 , … , u n ) u 1 , … , u n (u_(1),dots,u_(n))\left(\mathbf{u}_{1}, \ldots, \mathbf{u}_{n}\right)(u1,…,un) and ( v 1 , … , v n ) v 1 , … , v n (v_(1),dots,v_(n))\left(\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right)(v1,…,vn), and the determinant of both has the same sign, then both bases have the same orientation. For example, assume that in R 4 R 4 R^(4)\mathbb{R}^{4}R4 an ordered basis ( e 1 , e 2 , e 3 , e 4 ) e 1 , e 2 , e 3 , e 4 (e_(1),e_(2),e_(3),e_(4))\left(\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}, \mathbf{e}_{4}\right)(e1,e2,e3,e4) is given by e 1 = ( 1 , 3 , 2 , 0 ) , e 2 = ( 2 , 2 , 1 , 3 ) , e 3 = ( 3 , 1 , 3 , − 2 ) e 1 = ( 1 , 3 , 2 , 0 ) , e 2 = ( 2 , 2 , 1 , 3 ) , e 3 = ( 3 , 1 , 3 , − 2 ) e_(1)=(1,3,2,0),e_(2)=(2,2,1,3),e_(3)=(3,1,3,-2)\mathbf{e}_{1}=(1,3,2,0), \mathbf{e}_{2}=(2,2,1,3), \mathbf{e}_{3}=(3,1,3,-2)e1=(1,3,2,0),e2=(2,2,1,3),e3=(3,1,3,−2) and e 4 = ( 1 , − 1 , 3 , 3 ) e 4 = ( 1 , − 1 , 3 , 3 ) e_(4)=(1,-1,3,3)\mathbf{e}_{4}=(1,-1,3,3)e4=(1,−1,3,3). If we write these basis vectors out as a 4 × 4 4 × 4 4xx44 \times 44×4 matrix, we find the determinant to be -128 . Any other ordered basis in R 4 R 4 R^(4)\mathbb{R}^{4}R4 with a negative determinant also has the same orientation as ( e 1 , e 2 , e 3 , e 4 ) e 1 , e 2 , e 3 , e 4 (e_(1),e_(2),e_(3),e_(4))\left(\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}, \mathbf{e}_{4}\right)(e1,e2,e3,e4).
For any vector space (not just R n R n R^(n)\mathbb{R}^{n}Rn ), if we have two ordered bases E = ( u 1 , … , u n ) E = u 1 , … , u n E=(u_(1),dots,u_(n))E=\left(\mathbf{u}_{1}, \ldots, \mathbf{u}_{n}\right)E=(u1,…,un) and E ′ = ( v 1 , … , v n ) E ′ = v 1 , … , v n E^(')=(v_(1),dots,v_(n))E^{\prime}=\left(\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right)E′=(v1,…,vn) and a change-of-basis matrix A A AAA that transforms E ′ E ′ E^(')E^{\prime}E′ to E E EEE, we can write the matrix equation
E = A E ′ E = A E ′ E=AE^(')E=A E^{\prime}E=AE′
E E EEE and E ′ E ′ E^(')E^{\prime}E′ then have the same orientation if the determinant of A A AAA is positive.
Figure 6.2: Orientation of manifolds.
Moving on from orientations of vector spaces - how do we decide the orientation of a manifold? We won't delve into the mathematical details, but instead give a (very) basic overview. Figure 6.2 (from Penrose [17]) illustrates the orientations of four different manifolds. A 0 -manifold (0-dimensional manifold) is a set of discrete points.
An orientation on a 0 -manifold is a choice of sign, positive ( + ) ( + ) (+)(+)(+) or negative ( − ) ( − ) (-)(-)(−), for each point. The orientation of a 1-manifold (one-dimensional manifold), or curve, is a choice of direction by which we move along the curve. The orientation of a 2 -manifold can be illustrated using little circles showing the rotation of tangent vectors at a point in a 'positive' direction. In the words of Penrose: 'For a 3-manifold the orientation specifies which triad of independent vectors at a point is to be regarded as "right-handed" and which as "left-handed".
Recall that we earlier noted that the tangent space is vector space. Put simply, if all the tangent spaces on a manifold can be oriented consistently, we say the manifold is orientable. We've just seen that the orientation of an n n nnn-dimensional vector space can be understood in terms of the sign of an n × n n × n n xx nn \times nn×n determinant. We know that an n × n n × n n xx nn \times nn×n determinant is a multilinear alternating function of n n nnn vectors (written as columns or rows). We also know that a differential n n nnn-form is a linear or, more generally, multilinear alternating function of n n nnn tangent vectors. It shouldn't come as a great surprise, therefore, that differential n n nnn-forms are involved in the concept of manifold orientation. The idea, which at our level we don't use but is worth mentioning in passing, is that an n n nnn-dimensional manifold M M MMM is orientable if and only if there exists on M M MMM a nowhere-zero differential n n nnn-form. For example, the 2 -form d x ∧ d y d x ∧ d y dx^^dyd x \wedge d ydx∧dy is nowhere-zero on R 2 R 2 R^(2)\mathbb{R}^{2}R2, therefore R 2 R 2 R^(2)\mathbb{R}^{2}R2 is orientable. Fortney [9] states:
Thus the set of all nowhere-zero n n nnn-forms splits into two equivalence classes, one equivalence class that consists of all the everywhere positive nowherezero n n nnn-forms and a second equivalence class that consists of all the everywhere negative nowhere-zero n n nnn-forms. By choosing one of these two equivalence classes we are specifying what is called an orientation of the manifold.
Now let's return to earth and take a more practical look at orientation with reference to the elementary manifolds we are concerned with in this book. The orientation of a parameterised curve is usually taken to be in the direction of increasing values of the parameter. An ordinary two-sided surface in R 3 R 3 R^(3)\mathbb{R}^{3}R3 will, at every point, have two unit normal vectors n n n\mathbf{n}n and − n − n -n-\mathbf{n}−n, each pointing in opposite directions. The surface of a sphere, for example, will have unit normal vectors pointing inwards and outwards. We give the surface an orientation by choosing one of these two sets of unit normal vectors. Not all surfaces are orientable. The Möbius strip, for example, is a one-sided surface and therefore not orientable.
There are various ad hoc rules that reflect the standard right-handed orientation of R 3 R 3 R^(3)\mathbb{R}^{3}R3. The right-hand rule used for the cross-product, for example. Green's theorem only works (without a change of sign) when the boundary curve has a positive orientation if it is traced out in a counterclockwise direction. Stokes' theorem involves a surface S S SSS with boundary curve C C CCC. The orientation of S S SSS (determined by the choice of normal vector n ) n ) n)\mathbf{n})n) must match the orientation of the boundary C C CCC. The rule is that the direction of C C CCC is considered positive if you are walking around C C CCC with your head pointing in the direction of n n n\mathbf{n}n and the surface is on your left.
  • The good news as far as we're concerned is that almost all the manifolds we encounter in this book are described parametrically. When we parameterise
    a manifold M M MMM, that parameterisation will induce an orientation on M M MMM. This induced orientation is the one we'll be using.
Finally, one aspect of orientation that we must take on board with regard to differential forms is the necessity of keeping the differentials in the correct order (so that the sign comes out right). For example, the standard positive orientation built into R 2 R 2 R^(2)\mathbb{R}^{2}R2 means the correct order of d x , d y d x , d y dx,dyd x, d ydx,dy is
d x ∧ d y d x ∧ d y dx^^dyd x \wedge d ydx∧dy
And the standard right-handed orientation built into R 3 R 3 R^(3)\mathbb{R}^{3}R3 means the correct order of d x , d y d x , d y dx,dyd x, d ydx,dy and d z d z dzd zdz is
d x ∧ d y ∧ d z d x ∧ d y ∧ d z dx^^dy^^dzd x \wedge d y \wedge d zdx∧dy∧dz
Of course, we're not confined to R 3 R 3 R^(3)\mathbb{R}^{3}R3. Say we have an n n nnn-form in R n R n R^(n)\mathbb{R}^{n}Rn with coordinates u 1 , u 2 , … , u n u 1 , u 2 , … , u n u^(1),u^(2),dots,u^(n)u^{1}, u^{2}, \ldots, u^{n}u1,u2,…,un. By choosing an order for the coordinates, we choose an orientation for our R n R n R^(n)\mathbb{R}^{n}Rn space. If we choose the obvious order u 1 , u 2 , … , u n u 1 , u 2 , … , u n u^(1),u^(2),dots,u^(n)u^{1}, u^{2}, \ldots, u^{n}u1,u2,…,un, the correct order of d u 1 , d u 2 , … , d u n d u 1 , d u 2 , … , d u n du^(1),du^(2),dots,du^(n)d u^{1}, d u^{2}, \ldots, d u^{n}du1,du2,…,dun is straightforward:
d u 1 ∧ d u 2 ⋯ ∧ d u n d u 1 ∧ d u 2 ⋯ ∧ d u n du^(1)^^du^(2)cdots^^du^(n)d u^{1} \wedge d u^{2} \cdots \wedge d u^{n}du1∧du2⋯∧dun
Interchanging any two of the d u i d u i du^(i)d u^{i}dui differentials introduces a minus sign (due to the anti-commutativity of the wedge product) and therefore puts the thing in the wrong order. So, for example,
d u 2 ∧ d u 1 ⋯ ∧ d u n d u 2 ∧ d u 1 ⋯ ∧ d u n du^(2)^^du^(1)cdots^^du^(n)d u^{2} \wedge d u^{1} \cdots \wedge d u^{n}du2∧du1⋯∧dun
is in incorrect order, because
d u 2 ∧ d u 1 ⋯ ∧ d u n = − d u 1 ∧ d u 2 ⋯ ∧ d u n d u 2 ∧ d u 1 ⋯ ∧ d u n = − d u 1 ∧ d u 2 ⋯ ∧ d u n du^(2)^^du^(1)cdots^^du^(n)=-du^(1)^^du^(2)cdots^^du^(n)d u^{2} \wedge d u^{1} \cdots \wedge d u^{n}=-d u^{1} \wedge d u^{2} \cdots \wedge d u^{n}du2∧du1⋯∧dun=−du1∧du2⋯∧dun
Assuming it has one, an n n nnn-dimensional manifold M M MMM has a boundary ∂ M ∂ M del M\partial M∂M of dimension n − 1 n − 1 n-1n-1n−1. If you ever need to integrate over a boundary, you will need an n − 1 n − 1 n-1n-1n−1 differential form. It's therefore instructive to understand the correct order (with d u i d u i du^(i)d u^{i}dui omitted) for ( n − 1 ) ( n − 1 ) (n-1)(n-1)(n−1)-forms. This one isn't quite so straightforward. Here it is:
(6.0.1) ( − 1 ) ( i − 1 ) d u 1 ⋯ ∧ d u ( i − 1 ) ∧ d u ( i + 1 ) ⋯ ∧ d u n (6.0.1) ( − 1 ) ( i − 1 ) d u 1 ⋯ ∧ d u ( i − 1 ) ∧ d u ( i + 1 ) ⋯ ∧ d u n {:(6.0.1)(-1)^((i-1))du^(1)cdots^^du^((i-1))^^du^((i+1))cdots^^du^(n):}\begin{equation*} (-1)^{(i-1)} d u^{1} \cdots \wedge d u^{(i-1)} \wedge d u^{(i+1)} \cdots \wedge d u^{n} \tag{6.0.1} \end{equation*}(6.0.1)(−1)(i−1)du1⋯∧du(i−1)∧du(i+1)⋯∧dun
We'll do a quick example before justifying this formula.
Example 6.1. From Parkinson [15]. We have a 5 -form ω = d u 1 ∧ d u 2 ∧ d u 3 ∧ d u 4 ∧ d u 5 ω = d u 1 ∧ d u 2 ∧ d u 3 ∧ d u 4 ∧ d u 5 omega=du^(1)^^du^(2)^^du^(3)^^du^(4)^^du^(5)\omega=d u^{1} \wedge d u^{2} \wedge d u^{3} \wedge d u^{4} \wedge d u^{5}ω=du1∧du2∧du3∧du4∧du5 in R 5 R 5 R^(5)\mathbb{R}^{5}R5. What is the correct order for a 4 -form with (a) d u 2 d u 2 du^(2)d u^{2}du2 removed, and (b) d u 3 d u 3 du^(3)d u^{3}du3 removed. (a) d u 2 d u 2 du^(2)d u^{2}du2 means i = 2 i = 2 i=2i=2i=2, and (6.0.1) becomes
( − 1 ) ( 2 − 1 ) d u 1 ∧ d u 3 ∧ d u 4 ∧ d u 5 = ( − 1 ) 1 d u 1 ∧ d u 3 ∧ d u 4 ∧ d u 5 = − d u 1 ∧ d u 3 ∧ d u 4 ∧ d u 5 ( − 1 ) ( 2 − 1 ) d u 1 ∧ d u 3 ∧ d u 4 ∧ d u 5 = ( − 1 ) 1 d u 1 ∧ d u 3 ∧ d u 4 ∧ d u 5 = − d u 1 ∧ d u 3 ∧ d u 4 ∧ d u 5 {:[(-1)^((2-1))du^(1)^^du^(3)^^du^(4)^^du^(5)],[=(-1)^(1)du^(1)^^du^(3)^^du^(4)^^du^(5)],[=-du^(1)^^du^(3)^^du^(4)^^du^(5)]:}\begin{aligned} & (-1)^{(2-1)} d u^{1} \wedge d u^{3} \wedge d u^{4} \wedge d u^{5} \\ & =(-1)^{1} d u^{1} \wedge d u^{3} \wedge d u^{4} \wedge d u^{5} \\ & =-d u^{1} \wedge d u^{3} \wedge d u^{4} \wedge d u^{5} \end{aligned}(−1)(2−1)du1∧du3∧du4∧du5=(−1)1du1∧du3∧du4∧du5=−du1∧du3∧du4∧du5
(b) d u 3 d u 3 du^(3)d u^{3}du3 means i = 3 i = 3 i=3i=3i=3, and (6.0.1) becomes
( − 1 ) ( 3 − 1 ) d u 1 ∧ d u 2 ∧ d u 4 ∧ d u 5 = ( − 1 ) 2 d u 1 ∧ d u 2 ∧ d u 4 ∧ d u 5 = d u 1 ∧ d u 2 ∧ d u 4 ∧ d u 5 ( − 1 ) ( 3 − 1 ) d u 1 ∧ d u 2 ∧ d u 4 ∧ d u 5 = ( − 1 ) 2 d u 1 ∧ d u 2 ∧ d u 4 ∧ d u 5 = d u 1 ∧ d u 2 ∧ d u 4 ∧ d u 5 {:[(-1)^((3-1))du^(1)^^du^(2)^^du^(4)^^du^(5)],[=(-1)^(2)du^(1)^^du^(2)^^du^(4)^^du^(5)],[quad=du^(1)^^du^(2)^^du^(4)^^du^(5)]:}\begin{aligned} & (-1)^{(3-1)} d u^{1} \wedge d u^{2} \wedge d u^{4} \wedge d u^{5} \\ & =(-1)^{2} d u^{1} \wedge d u^{2} \wedge d u^{4} \wedge d u^{5} \\ & \quad=d u^{1} \wedge d u^{2} \wedge d u^{4} \wedge d u^{5} \end{aligned}(−1)(3−1)du1∧du2∧du4∧du5=(−1)2du1∧du2∧du4∧du5=du1∧du2∧du4∧du5
We can understand (6.0.1) by noting that
d u i ∧ ( − 1 ) ( i − 1 ) d u 1 ⋯ ∧ d u ( i − 1 ) ∧ d u ( i + 1 ) ⋯ ∧ d u n = d u 1 ∧ d u 2 ⋯ ∧ d u n d u i ∧ ( − 1 ) ( i − 1 ) d u 1 ⋯ ∧ d u ( i − 1 ) ∧ d u ( i + 1 ) ⋯ ∧ d u n = d u 1 ∧ d u 2 ⋯ ∧ d u n du^(i)^^(-1)^((i-1))du^(1)cdots^^du^((i-1))^^du^((i+1))cdots^^du^(n)=du^(1)^^du^(2)cdots^^du^(n)d u^{i} \wedge(-1)^{(i-1)} d u^{1} \cdots \wedge d u^{(i-1)} \wedge d u^{(i+1)} \cdots \wedge d u^{n}=d u^{1} \wedge d u^{2} \cdots \wedge d u^{n}dui∧(−1)(i−1)du1⋯∧du(i−1)∧du(i+1)⋯∧dun=du1∧du2⋯∧dun
because the d u i d u i du^(i)d u^{i}dui has to jump over i − 1 i − 1 i-1i-1i−1 differentials until it gets back to its rightful place between d u ( i − 1 ) d u ( i − 1 ) du^((i-1))d u^{(i-1)}du(i−1) and d u ( i + 1 ) d u ( i + 1 ) du^((i+1))d u^{(i+1)}du(i+1), with each jump changing the sign of the expression.
Returning to R 3 R 3 R^(3)\mathbb{R}^{3}R3, the correct order for the basis 2 -forms is
d y ∧ d z , − d x ∧ d z , d x ∧ d y d y ∧ d z , − d x ∧ d z , d x ∧ d y dy^^dz,-dx^^dz,dx^^dyd y \wedge d z,-d x \wedge d z, d x \wedge d ydy∧dz,−dx∧dz,dx∧dy
which we can rewrite as
d y ∧ d z , d z ∧ d x , d x ∧ d y d y ∧ d z , d z ∧ d x , d x ∧ d y dy^^dz,dz^^dx,dx^^dyd y \wedge d z, d z \wedge d x, d x \wedge d ydy∧dz,dz∧dx,dx∧dy
  1. 1 1 ^(1){ }^{1}1 Shameless plug - the result of my labours was a book: A Most Incomprehensible Thing: Notes Towards a Very Gentle Introduction to the Mathematics of Relativity (ISBN 9780957389465).
  2. 1 1 ^(1){ }^{1}1 We are here using the Einstein summation convention, meaning that if a single term contains the same upper and lower index ( i i iii in this case), a sum is implied.
  3. 2 A 2 A ^(2)A{ }^{2} \mathrm{~A}2 A vector space V V VVV is a group of objects (called vectors) that (a) include a zero vector, (b) include an inverse vector, and (c) can be added together, and multiplied by numbers according to a specific set of rules (the vector space axioms). The result of these operations is another member of V V VVV. R n R n R^(n)\mathbb{R}^{n}Rn, for example, is a vector space. Consider the Euclidean plane R 2 R 2 R^(2)\mathbb{R}^{2}R2, where a vector is defined as a pair of real numbers ( x , y ) ( x , y ) (x,y)(x, y)(x,y). If we add two vectors ( x 1 , y 1 ) x 1 , y 1 (x^(1),y^(1))\left(x^{1}, y^{1}\right)(x1,y1) and ( x 2 , y 2 ) x 2 , y 2 (x^(2),y^(2))\left(x^{2}, y^{2}\right)(x2,y2), we get another pair of real numbers ( x 1 + x 2 , y 1 + y 2 ) x 1 + x 2 , y 1 + y 2 (x^(1)+x^(2),y^(1)+y^(2))\left(x^{1}+x^{2}, y^{1}+y^{2}\right)(x1+x2,y1+y2), ie another vector. And if we multiply a vector ( x , y ) ( x , y ) (x,y)(x, y)(x,y) by a scalar (ie a number) k k kkk we get ( k x , k y ) ( k x , k y ) (kx,ky)(k x, k y)(kx,ky), which is also a vector. A basis for V V VVV is a set of linearly
  4. 1 1 ^(1){ }^{1}1 We are here using the Einstein summation convention, meaning that if a single term ( ∂ x i ∂ y j ∂ ∂ x i ) ∂ x i ∂ y j ∂ ∂ x i ((delx^(i))/(dely^(j))(del)/(delx^(i)))\left(\frac{\partial x^{i}}{\partial y^{j}} \frac{\partial}{\partial x^{i}}\right)(∂xi∂yj∂∂xi) contains the same upper and lower index ( i i iii in this case), a sum is implied.